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Photographic 

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CIHM/ICMH 

Microfiche 

Series. 


CIHM/ICMH 
Collection  de 
microfiches. 


Canaf^ian  Institute  for  Historical  Microreproductions  Institut  canadien  de  microreproductions  historiques 

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^ 


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10X 

14X 

18X 

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/ 

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e 

Stalls 
s  du 
lodifier 
r  une 
Image 


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3S 


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et  de  haut  en  bas,  en  prenant  le  nombre 
d'images  ndcessaire.  Les  diagrammes  suivants 
illustrent  la  m^thode. 


errata 

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2 

3 

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5 

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1] 

I 


SELF-INSTRUCTOR 


r 


\ 


ON 


LUMBER  SURVEYING 


FOR  THE  USE  OP 


t 


LUMBER  MANUFACTURERS,  SURVEYORS, 
AND  TEACHERS. 


BY 


CHARLES  KIlSrSLET, 

PRACTICAL  SURVEYOR  AND  TEACHER  OF  SURVEYING. 


ASSIGNOR, 

JAMES   KINSLEY. 


PUBLISHED  BY  THE  AUTHOR. 

Calais,  Me.,  and  St.  Stephen,  N.  B. 

1870. 


f/ 


^P'feS^/A  ^*;..:4^^  ^^,^',Jr/f.^J' .  ^ 


ErVERSIBE,   cambridob: 

ELECTROTYPED    AND    PRINTED    BT 
H.   0.   HOUGaiON  AND  COMPANY. 


PREFACE. 


This  work  combines  the  theoretical  and  practical 
parts  of  surveying,  in  such  a  manner  as  to  enable  the 
energetic  and  uninitiated  student  who  applies  himself 
to  the  study  of  this  useful  and  interesting  science  for  a 
short  time,  to  survey  all  kinds  of  lumber  with  accu- 
racy  and  expertness.  It  contains  tables  for  measuring 
boards,  plank,  deal,  and  timber  by  board  measure, 
by  which  the  Surveyor  can  dispense  with  the  use  of 
the  Board  Rule.  It  contains  the  rules  generally 
adopted  by  Surveyors,  and  also  a  more  concise  rule 
than  that  in  general  use  :  for  plank,  deal,  and  timber, 
this  rule  alone  is  worth  more  than  the  price  of  the 
book  to  any  Surveyor,  as  it  requires  less  mental  calcu- 
lation than  by  the  other  rules,  enabling  him  to  sur- 
vey flistcr  and  with  less  trouble  than  he  could  other- 
wise do.  It  contains  tables  for  incli,  incli  and  a 
quarter,  and  inch  and  a  half  boards  for  battens  and 
joist.  It  also  contains  rules  and  tables  for  surveying 
logs  by  board  and  cubic  measure,  and  rules  for  ton  tim- 
ber. It  also  contains  tables  showing  the  number 
of  feet  in  length,  of  any  dimension,  which  will  make 
1,000  feet  board  measure  or  1,000  feet  cubic  measure; 


IV 


PREFACE. 


a  new  method  of  finding  the  soHd  contents  of  timber  ; 
a  rule  for  finding  what  a  round  log  will  square,  by  hav- 
ing the  circumference  or  diameter  given,  or  in  other 
words,  to  find  the  inscribed  square  ;  how  to  make  out 
specifications,  survey  bills,  etc. ;  rule  for  measuring 
tapering  timber ;  table  of  quarter-girts  for  logs  ;  rule 
for  finding  how  much  in  length,  of  any  dimension, 
which  will  make  a  solid  foot,  or  any  otlier  desired 
quantity ;  table  showing  the  weight  of  twenty-five 
kinds  of  wood,  with  a  rule  for  finding  the  weight  of 
the  same  from  the  contents ;  the  English  and  Amer- 
ican Government  rules  for  finding;  the  tonnage  of  ves- 
sels,  and  rules  for  gaun-ino;  and  ullamno;  casks.  It  also 
contains  a  correct  and  extensive  interest  table. 


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i 


TABLE  OF   CONTENTS. 


om  Two 


Rule  for  measuring  Rectangular  Boards         .       . 

Exercise  on  the  Rule 

To  find  the  Contents  of  a  Triangular  Board  . 

Exercise  on  the  Rule 

To  iind  the  Solidity  of  a  Sphere  or  Globe 

To  find  the  Contents  of  a  Triangular  Solid 

To  find  the  Superficial  Contents  of  a  Globe    . 

To  find  the  Contents  of  a  Circular  Board    .... 

Rule  for  finding  the  Contents  of  a  Circle 

Rule  for  finding  the  Diameter  or  Circumference  of  a  Circle 

Table  for  measuring  Inch  Boards,  without  a  Board  Rule,  fi 

Inches  to  Thirty-six  Inches  wide      .... 

Exercise  on  the  Table 

Table  for  measuring  Inch-and-a-Quarter  Boards  without  a  Board  Rule 

Exercise  on  the  Table 

Table  for  Inch-and-a-IIalf  Boards  without  using  the  Board  Rule  . 

Exercise  oi.  the  Rule 

Table  for  Plank  from  Two  to  Thirty  Inches  wide  .... 

Exercise  on  the  Table 

Table  for  Three-inch  Deals  from  Three  to  Twenty-four  Inches  wide 
Table  for  Four  inch  Deals  from  Four  to  Twelve  Inches  wide 
Table  for  Five-inch  Timber  from  Five  to  Twelve  Inches  wide 
Table  for  Six-inch  Timber  from  Six  to  Twelve  Inches  wide 
Table  for  Seven-inch  Timber  from  Seven  to  Twelve  Inches  wide  . 
Table  for  Fight-inch  Timber  from  Eight  to  Twelve  Inches  wide 
Table  for  Nine-inch  Timber  from  Nine  to  Twelve  Inches  wide 
Table  for  Ten-inch  Tiuil)er  from  Ten  to  Twelve  Inches  wide 
Table  for  Eleven-inch  Timber  from  Eleven  to  Twelve  Inches  wide 
Table  for  Twelve-inch  Timber  from  Twelve  to  Twenty  Inches  wide  . 
Plan  of  drawing  a  Plank  Shingle,  with  Directions  for  dotting  Plank 

Deal,  &c 

Rule  for  Plank  Specifications 

How  to  keep  a  Joist  or  Scantling  Shingle       .        .        . 
New  York  Deals,  Three-inch       ....        * 

Specification  Rule  for  Three-inch 

Specification  Rules  for  Four-inch         .... 
Five-inch  Timber  Shingle,  with  Rule  for  Specification  . 


Page 
7 
7 
8 
8 
9 
9 
9 
9 
9 
9 

10 
10 
11 
11 
12 
12 
13 
13 
14 
15 
16 
16 
17 
17 
17 
17 
17 
17 

18,19 
20 
22 
23 
24 
25,26 
28-30 


VI 


TABLE  OF  CONTENTS. 


Rule  for  Six-iiicli  Specification 31 

Ilulc  for  Seven-inch  Specillcution 33 

Kule  Tor  Kiylit-incli  Speciticution 36 

Kule  lor  Nine-incii  Specilication 37 

Kule  (or  i'en-incli  Specilicafioii 40 

Kule  for  Kleven-inch  Specification 41 

Kule  for  Twelve  inch  Specilication 42 

Kule  for  lindiui,'  the  Contents  of  Hattens  or  Two-anda-Half-inch  Stuff  44 

SpHcilicatioii  of  Hatten  Shinj^Ie  and  Kule 45 

Kandoni  Shingle,  Contents  given  in  the  Columns 45 

Kan<lon>  Shingle,  Kunning  Lengths  in  the  Columns   ....  46 
Table  showing  the  Number  of  Feet  in  Length,  of  all  Dimensions,  that 

will  make  1,000  teet  of  Boanl  Measure 47 

Table  showing  the  Number  of  Feet  in  Length,  of  all  Dimensions,  from 
Five  Inches  by  Five  Inches  to  Twenty-two  Inches  by  Twenty- 
Four  Inches  that  will  make  1,000  Cubic  teet,  with  Kules  showing 

how  both  Tables  are  computed 48 

New  Rules  (or  finding  the  Contents  in  Cubic  Feet  of  Timber,  from  Five 

by  Five  up 49 

Second  Method  of  making  out  Specification,  and  Rule       .        .        .51,  52 

Specification  of  I'liiladelphia  Deal,  and  Rule 53,  54 

How  to  use  the  Board  K'ule,  with  Kxercise 54 

Rule  for  measuring  Logs,  with  Example 55 

To  find  the  Largest  Square  Piece  of  Timber  that  can  be  sawed  from 

a  Round  Log,  by  having  the  Circumference  or  Diameter  given  .  56 

Form  of  Bills  of  Lading 57,58 

Surveyor's  Hills  and  Receipts 59 

New  Rules  for  finding  the  Superficial  Contents  of  Plank,  Deal,  Joist, 

Battens,  and  Timber 59 

Given  the  Dimensions  of  the  End  of  a  Plank  to  find  what  Length  of 

it  will  make  a  Foot 61 

To  find  the  Solid  Contents  of  a  Piece  of  Tapering  Timber    .        .        .61 
When  a  Boarl  or  Plank  is  wider  at  one  End  than  the  other,  to  find 

■what  Length  of  it  will  make  a  Foot,  or  any  Desired  Quantity    .  62 
To  find  how  much  in  Length  will  make  a  Solid  Foot,  or  any  otiier  De- 
sired (Quantity,  of  Squared  Timber  of  Equal  Dimensions  from  End 

to  End 62 

Table  for  measuring  Round  Timber  by  the  Qnarter-girt  Areas  .        .  C3 
Table  and  Rule  for  finding  the  Weight  of  T'imber  from  a  Survey  of  its 

Contents 64 

English  Government  Rule  for  finding  the  Tonnage  of  Vessels  .        .  65 

United  States  Government  rule  for  finding  the  Tonnage  of  Vessels       .  66 

Gauging  of  Casks 67 

Ullaging  of  Casks 68 

Questions  (or  Exercise 69 

Log  Rule  for  Round  Timber 72-78 

Directions  for  \ising  the  Log  Rule        .......  77 

Interest  Table  and  Rule 77-79 


SELF-TNSTRUCTOR 

OM 

LUMBEE   SUEYEYma 


Rule  for  measuring  RectariQular  Boards. 

Multiply  the  length  in  feet  by  the  width  in  inches,  and 
divide  the  product  by  12,  to  find  the  contents  in  superficial 
feet.  Or  multiply  the  length  in  inches  by  the  width  in 
inches,  and  divide  by  144,  the  number  of  inches  in  a  square 
foot,  for  the  contents  in  superficial  feet. 

P.  S.  —  A  Rectangle  is  a  plain  figure  bounded  by  four 
straight  lines,  which  are  equal  and  pandlel,  and  Avhose 
angles    are    right    angles, 


as  B. 


B. 


QUESTIONS    FOR    EXERCISE. 

1.  What  are  the  contents  in  feet  of  a  rectangular  board 
30  feet  long  and  20  inches  wide  ?  Ans.  50  feet. 

2.  How  many  feet  in  a  board  26  feet  6  inches  long,  12 
inches  in  width  ?  Ans.  26^  feet. 

3.  AVhat  will  be  the  cost  of  a  walnut  board  32  feet  long 
and  16  inches  wide,  at  8  cents  per  square  foot.     Ans.  $3.41. 

4.  What  are  the  contents  of  a  board  22  feet  8  inches  long, 
and  1  foot  9  inches  in  width  ?  Ans.  39  feet  8  inches. 

When  a  Board  is  wider  at  one  End  than  at  the  other, 

Jlule.  —  Add  the  width  of  both  ends  together,  and  take 
half  the  sum  for  a  mean  width,  and  multiply  the  width  thus 
found  by  the  length,  for  the  contents  ;  or  take  the  width  in 


^ 


fr 


8 


SELF-INSTRUCTOR 


the  middle  of  the  board  and  multiply  by  the  length,  for  the 
contents. 

EXAMPLE. 

1.  What  are  the  contents  of  a  board  14  inshes  at  one  end 
and  20  inches  at  the  other,  and  24  feet  in  length. 

Ans.  34  feet. 
14  -|"  20  =  34  -i-  2  =  17,  mean  width  in  inches,  which 
multiplied  by  the  length,  24  feet  =  408;  408  4-  12  =  34 
feet  =  contents. 

2.  What  are  the  contents  of  a  board  2G  feet  long,  which 
measures  IG  inches  in  the  middle  ?        Ans.  34  feet  8  inches. 

26  feet  X  10  =  416 ;  416  h-  12  =  34  feet  8  mches  = 
contents. 

To  jind  the  Contents  of  a  Triangular  Board. 

Ride.  —  Multiply  the  length  in  feet  by  the  width  in 
inches,  and  take  half  the  sum  for  the  contents  in  inches, 
which  being  divided  by  12  will  give  the  contents  in  feet  of 
board  measure. 

EXAMPLE. 

1.  What  are  the  contents  of  the  board 
ABC,  whose  base  B  C  is  26  inches, 
and  perpendicular  height  A  D  is  18 
feet.  Ans.  19  feet  6  inches. 

18  X  26  =  468  -^  ^  =  234  -^  12 
=  10  feet  6  inches. 

2.  Wiiat  are  the  contents  of  the  trian- 
gular board  ABC,  whose  base  B  C  is 
2  feet  6  inches,  and  perpendicular  A  C,  24 
feet.  Ans.  30  feet. 

24  feet  X  2i  =  60  feet ;  60  feet  -^  2 
=  30  feet.    Or"— 

2  feet  6  inches  =  30  inches ;  30  inches 
X  24  feet  =  720  inches ;  720  ~  2  =  360 
inches  =  contents;  360  -r  12  =  30  feet's 
=  contents  in  feet. 


Ill 


ON  LUMBER  SURVEYING. 


9 


E 


The  contents  of  a  triangulnr  solid  can  be  found  in  the 
same  manner  by  the  foregoing  rule,  by  multiplying  the  con- 
tents thus  found  by  the  thickness  of  the  solid. 

How  many  feet  of  boards  in  a  triangular 
piece  of  timber,  A  B  C,  whose  length  A 
B  is  24  feet,  breadth  B  C  18  inches,  and 
thickness  C  E  2  feet  6  inches  ? 

24  feet  X  18  inches  =  432 ;    432-^-2 
=  216  inches;    21G  inches  -J-  12=18 
feet  =  contents  of  .sui)erfK'ial  triangle  A  B 
C,  which  being  multiplied  by  the  thickness, 
C  E,  2  feet  6  inches,  will  give  the  contents^ 
of  the  solid  triangle  A  B  C  D  E  F,  18  feet  X  H  feetc= 
Ans.  45  cubic  feet,  or  540  board  measure. 

For  Measurement  of  a   Globe. 

Rule.  —  To  find  the  solidity  of  a  globe,  cube  the  diame- 
ter, and  multiply  the  product  by  5,23G ;  and  to  find  the  sur- 
face of  a  globe,  multiply  the  diameter 
by  the  circumference.  To  find  the  cir- 
cumference by  having  the  diameter 
given,  say  as  7  is  to  22,  so  is  the  diame- 
ter to  the  circumference,  or  as  22  is  to  7, 
so  is  the  circumference  to  the  diameter. 

To  find  the  Contents  of  a  Circle. 

Mule  1.  —  ]\Iultiply  half  the  circum- 
ference by  half  the  diameter,  for  tlio  contents. 

Mule  2.  —  Square  the  diameter,  and  multiply  it  by  .7854 
for  the  contents,  or  square  the  circumference,  and  multiply  it 
by  .07958  for  the  contents. 

P.  S.  —  The  square  of  a  number  is  found  by  multiplying 
the  number  by  itself.  r  j    ts 


10 


SELF-INSTRUCTOR 


I 


Table  for  measuring  Inch  Boards  without  a  Rule,  from 
2  Inches  to  '66  Inches  wide. 


ii 


Inches.       Feet. 

Inche.s.         Feet. 

Inches.         Feet. 

Indies.         Feet. 

2X1—^ 

11X1      H 

20  X  1  —  if 

29  X  1        2/'o 

3X1        i 

12X  1         1 

21X1        1^ 

30  XI  — 2^ 

4X1        ^ 

13X1  — irV 

22  X  1        1^ 

31  X  1        2V^ 

5X1  — A 

14X  1  —  1^ 

23  X  1        Hh 

32  X  1        2^ 

6X  1        i 

lb  XI       li 

24  X  1  — 2 

33  X  1        2^ 

7  X  1        tV 

16  X  1         H 

25  X  1         2iL 

34  X  1  —  2if 

8Xl-§ 

17  X  1         1t\ 

26  X  1         2^ 

35X1        2i.V 

9X1     i 

18  X  1  — U 

27  X  1        2^ 

36  X  1        3 

10X1      ^ 

19X1         h^j 

28  X  1        2i 

In  order  to  niirvey  boards  by  the  Tuble  of  Board  IMeas- 

ure,  the  Surveyor  must  commit  the  table  to  m(Mnory,  and 

by  a  little  practice,  he  will  become  expert  at  surveying  by 
this  method. 

Questions  for  Exercise  done  hj  the   Table  of  Board 

Measure. 

1.  What  are  thf.  contents  of  a  board  24  feet  long  and  18 
inches  wide  ?  Ans.  2  4  X  U  —  36  teet. 

2.  How  many  feet  in  a  board  32  feet  long  and  17  inches 
wide  ?  Ans.  45^  feet. 

By  the  table,  17  inches  wide  is  ly*^  the  length,  for  the 
contents;  therefore  32  feet  X  IfV  =  ^^3  ^^^^' 

3.  What  are  the  contents  of  a  board  21  feet  6  inches  long 
and  6  inches  wide  ?  Ans.  10  feet  9  inches. 

By  tiie  table,  6  inches  wide  is  half  the  length,  for  the  con- 
tents;  therefore  21  feet  6  inches  -f-  2  =  10  feet  9  niches  = 
contents. 

4.  Required  the  contents  of  a  board  36  feet  long  and  3 
inches  wide  ?  Ans.  3G  -j-  4  =:  9  feet. 

5.  Find  the  contents  of  a  board  24  feet  8  inches  long  and 
14  inches  wide  ? 

Ans.  24  feet  8  inches  X  1 J  =  28  feet  9  inches  4'\ 


i    t 


ON   LUMBER   SURVEYING. 


11 


6.  Required  the  contents  of  a  board  27  feet  long  and  30 
inches  wide  ?  Ans.  67^  feet. 

7.  What  is  the  value  of  a  walnut  board  23  feet  6  inches 
long,  and  36  inches  wide,  @  12|^  cents  per  square  foot? 

Ans.  $8.8 li 

8.  Required  the  contents  of  a  board  16  feet  long  and  "7 
inches  wide  ?  Ans.  36  feet. 

9.  How  many  feet  in  a  board  38  feet  long  and  28  inches 
wide  ?  Ans.  88  feet  8  inches. 

10.  Required  the  contents  of  a  board  16  feet  long  and  19 
inches  in  width  ?  Ans.  25  feet  4  inches. 


Table  for  Inch-and-a- Quarter   Boards,  from   2    Indies   to 

36  Inches  wide. 


Tnches.          Feet. 

Inches.          Feet. 

Inches.          Feet. 

2xii  — A 

14Xli  =  lii 

26  X  Ir       ni 

3  X  li      tV 

i5Xii  — iyV 

27Xli  — 2}^ 

4  X  li      tV 

icxii  — i! 

28  X  I4 — 2}  I: 

5X1]    n 

i7Xii    i:^^- 

29  X  I4        3:jV 

6Xli  — f 

isxii  — 1# 

30  X  li        3i 

7xii    ri 

19Xli        115 

31  X  I4         3^4 

8Xii     # 

20  X  li  — 2i'.T 

32  X  li:^3^ 

9Xli        n 

2lXli  — 2/5 

33  X  li  -  3-rV 

10  X  li     1^^ 

22  X  li        2.,7j. 

34Xli        3^i 

iiXii  — 1t\ 

23Xli       2]| 

35Xli  — 3^^ 

i2Xii     li 

13Xli  — Ui 

24Xli  — 2^ 

36Xli  — 3| 

25xii    m 

Examples   of  \\-inch  Board   3feasure  done    ly  the   Table. 

1.  "What  are  the  contents  of  a  board  li  inches  thick,  32 
inches  wide,  and  30  feet  long?  Ans.  100  feet. 

By  the  table  32  inches  is  3^  times  the  length  ;  for  the  con- 
tents, therefore,  30  feet  X  3^  =  100  feet. 

2.  What  are  the  contents  of  a  board  li  inches  by  18  inches, 
and  36  feet  in  length  ?  Ans.  G7  feet  6  inches. 


X 


N 


r 


12 


SELF-INSTRUCTOR 


3.  Hequired  the  contents   of  a  board  1^   inches  by  24 
inches,  and  32  feet  8  inches  in  length  ? 

Ans.  81  feet  8  inches. 

4.  How  many  feet  in  a   1^-iuch  board  16  inches  wide 
and  24  feet  long  ?  Ans.  40  feet. 

5.  What  will  be  the  cost  of  a  piece  of  mahogany  1^  inches 
by  12  inches,  and  36  feet  long,  @  6  cents  per  foot  ? 

Ans.  $2.70. 

Table  for    One-and-a- Half-inch    Boards^  from   2   to   24 

Inches  wide. 


Inches.        Feet. 

Inches.            Feet. 

Inches.           Feet. 

Inches.         Feet. 

2Xii     i 

8X1^        1 

14X1^        1^ 

20  X  U        2i 

3  X  li        ^ 

9X1^  —  1)^ 

15X1^        1| 

21  X  li        2t 

4X1^        h 

10  X  U      ij 

16  X  li  — 2 

22X1^  — 2| 

5XU        t 

11  X  li      1^ 

17  X  1*        2i 

23  X  li        2^ 

6X1^    i 

12XU-  — li 

isxii— 2i 

24  X  1^  —  3* 

7X1^  —  ^ 

13  X  li        if 

19  X  li        2t 

Hi 


1.  What  are  the  contents  of  a  1^-inch  board  32  feet  long 
and  24  inches  wide  ?  Ans.  32  feet  X  3  feet=  96  feet. 

2.  Required  the  contents  of  a  1^-inch  board  18  feet  long 
and  18  inches  wide  ?  .    Ans.  40i-  feet. 

3.  Find  the  contents  of  a  board  1^  X  10  inches  and  28 
feet  8  inches  in  length  ?  Ans.  35  feet  10  inches. 

By  the  table  H  X  10  is  1;^  the  length,  for  the  contents. 
28  feet  8  inches  X  1^  =  35  feet  10  inches. 

4.  What  are  tlie  contents  of  a  board  24  feet  long,  20 
inches  wide,  and  H  inches  thick  ?  Ans.  60  feet. 

5.  Required  the  contents  of  a  board  16  inclies  wide,  1^ 
inches  thick,  and  27  feet  long.  Ans.  54  feet 

6.  What  is  the  value  of  a  board  17  inches  wide,  and  1^ 
inches  thick,  and  20  feet  long,  at  6  cents  per  foot  ? 

Ans.  $2.55. 

•  Equal  three  times  the  length,  for  contents. 


>  ! 


ON  LUMBER   SURVEYING. 


13 


s  by  24 

8  inches. 

les  wide 

40  feet. 

^  inches 

J.  $2.70. 

to   24 


s. 

Feet. 

4= 

2i 

li 

2i^ 

li 

2^ 

H 

2^ 

1* 

=  3* 

'eet  long 
96  feet, 
'eet  long 
lOi  feet, 
and  28 
inches, 
3ontents. 

ong,  20 

GO  feet. 

kvide,  1^ 

54  feet 

and  1^ 

.  $2.55. 


Table  for  Two-inch  or  Plank,  from  2  to  30  Inches  wide. 


Inches.        Feet. 

Inches.         Feet. 

Inches.         Feet. 

Inches.        Feet. 

2X2        ^ 
2X3  — i 

2  X  10  — if 

2  X  17        2^ 

2  X  24        4 

2X  11  —  1^ 

2X18  —  3 

2  X  23  —  4^ 

2X4-^ 

2X12        2 

2  X  19        3^ 

2  X  26        4\ 

2X3— if 

2  X  13        2^ 

2  X  20  —  3^ 

2  X  27        4h 

2X6        1 

2  X  14        2i 

2  X  21        3:^ 

2  X  28        4§ 

2X7        1^ 

2X  15  —  2^ 

2  X  22  —  3| 

2X29  —  4^ 

2X8     q 

2X9  — l| 

2X16        2f 

2X23_3g 

2  X  30        5 

WfL-m. 

EXERCISE. 


1.  Required  the  contents  of  a  plank  18  feet  long  and  15 


inches  in  width  ? 


Ans.  45  feet. 


9i 


times  the 
therefore 


By  the  table  15  inches  wide 
contents  in  feet  of  board  met 
=  45  feet. 

2.  Required  the  contents  of  a  plank  36  feet  long  and  12 
inches  wide  at  one  end,  and  16  inches  at  the  other  end? 

A71S.  84  feet. 
12  inches  -|-  16  inches  =  28  inches  ;  28  inches  -^  2  = 
mean  width  14  inches.     ]5y  the  table  14  inches  is  2^  times 
the  length  ;   therefore  36  feet  X  2^- =  84  feet. 

3.  What  is  the  value  of  a  plank  24  feet  long  and  27 
inches  wide  @  3^  cents  per  foot  ?  Ans.  $3.92. 

4.  Required  the  contents  of  a  plank  18  feet  long  and  4 
inches  wide  ?  Ans.  ^^  X  f  =  'V  =  ^2  feet. 

5.  "What  are  the  contents  of  1,860  feet  ruiuiing  lengths 
of  2  inches  X  2  inches  ?  Ans.  620  feet. 

Solution.  — 1  MO -^^=G20  feet. 

6.  In  2,500  feet  nmning  lengths  how  many  feet  contents 
of  2  inches  X  12  inches  ?  Ans.  5,000  feet  or  5  M.. 

2,500  feet  X  2  =  5,000  feet,  or  5  M. 


"''r* 


14 


SELF-INSTRUCTOR 


Table  for  Three-inch  Deals,  from  3  to  24  inches  wide. 

Inches.     Feet. 

Inches.        Feet. 

Inches.         Feet. 

Inches.        Feet. 

3X3  —  1 

3X    9        2^ 
3  X  10  — 2  J 

3X  15  —  3^ 

3  X  20  —  5 

3X4         1 

3  X  16  —  4 

3  X  21        5i 
3  X  22  —  bl 

3X5        li 
3X6        4 

3X11        2| 

3X  17  — 4^ 
3  X  18        4} 

3  X  12  —  3 

3  X  23  —  5-| 

3X7        1^ 

3  X  13=:  3^ 

3X  19  — 4f 

3X  24        6 

3X8        2 

3  X  14  —  32 

••' 


EXERCISE. 

1.  What  are  the  contents  of  a  deal  3  inches  thick,  6 
inches  wide,  ann  30  feet  long  ?  Ans.  45  feet. 

By  the  table  3X6  is  1|-  times  the  length,  for  the  con- 
tents ;  therefore  30  feet  X  H  =  45  =  contents. 

2.  What  are  the  contents  of  a  deal  3  inches  X  3  2  inches 
and  33^  feet  long?  Ans.  100  feet. 

3.  In  2,700  feet  of  running  lengths  of  3  inches  X  20 
inches,  how  many  feet  ?  Ans.  13,500  feet. 

By  the  table  3  X  20  is  5  times  the  length,  for  the  con- 
tents ;  2,700  X  5  =  13,500  feet. 

4.  Required  the  number  of  feet  running  lengths  of  3  X  4 
that  will  be  equal  to  2,000  feet  running  lengths  of  3  inches 
X  10  inches  ?  Ans.  5,000  feet. 

o.  What  number  of  feet  of  running  lengths  of  2  X  3 
will  be  equivalent  to  24,000  feet  running  lengths  of  3  X  12 
inches.  Ans.  144,000  feet. 

Solution.  —  By  the  tab'e  3  X  12  is  3  times  the  length,  for 
the  contents ;  therefore  24,000  feet  X  3  =  72,000  feet  = 
contents  of  3  X  12  inches,  and  by  the  table  2  X  3  is  =  to 
half  the  length,  for  the  contents ;  therefore  2  X  3  is  2  times 
the  contents  for  the  running  lengths,  consequently  72,000 
feet  X  2=  144,000  feet  running  length. 


ON  LUMBER  SURVEYING.  15 

Tahh  for  Four-inch  Deals,  from  A  to  12  Inches  wide. 


iches. 

Feet. 

X20: 

—  5 

X21: 

-51 

X22: 

X23: 

-5ii 

X24: 

=  6 

Inches.      Feet. 

Inches. 

Feet. 

Inches.         Feet. 

Inched.      Feet. 

4X4        1^ 

4X7: 

=  2^ 

4X    9        3 

4X11         3§ 

4X5        1§ 

4X8: 

24 

4  X  10        3  j 

4X  12        4 

4XC        2 

.EXKRCISi:. 

1.  "What  are  the  contents  of  a  deal  4X4  inches,  and  20 
feet  long  ?  Ans.  2(i|  feet. 

2.  What  are  the  contents  of  a  deal  4X5  and  24  i'eet 
long  ?  Ans.  40  i'eet. 

3.  Required  the  contents  of  a  deal  4  X  G  and  2(>  feet 
long?  Ans.  52  feet. 

4.  Required  the  contents  of  a  deal  4  niches  X  12  inches 
and  SO  feet  long  ?  Ans.  120  feet. 

5.  What  is  the  value  of  a  piece  of  oak  3G  feet  long,  4 
inches  thick,  and  11  inches  wide,  @  4^  cents  per  square 
ioot  ? 

6.  In  2,800  feet  of  running  lengths  of  4  inches  X  12 
inches,  how  many  feet  of  superficial  measurement  are  there? 

Ans.  11,200  feet. 

7.  How  many  feet  running  lengths  of  4  inches  X  12 
inches  deals  are  equal  to  3,000  feet  running  lengths  of  2  X 
6  ?  Ans.  750  feet. 

8.  What  is  the  amount  of  lumber  in  the  following  cargo, 
and  its  value  @  $15.00  per  M  ? 

Surveyed  from  Bennett  &  Co.,  of  Boston,  Mass.,  to  Ship 
Aurora,  Capt.  Jones,  — 

2,758  pieces  2  X  8  and  16  feet  long. 
3,800  pieces  4  X  12  and  30  feet  long. 
2,600  pieces  4  X  10  and  16  feet  long. 
250  M  of  Mer.  spruce  laths  @  $2.50  per  M. 

Ans.  653,41)7  feet  of  hnnber.     250  M  laths. 

Value  of  lumber,  $9,802.45^ 

Value  of  hitiis,        625.00*" 


$10,427. 45i- 


prr-^^' 


16 


SELF-INSTRUCTOR 


Table  of  Five-inch   Timher,  from  6  to  12  Inches  wide. 


Inches.        Feet, 
5X8  =  3^ 


Inches.        Feet. 
5X    9  =  3^ 
5  X  10  =  4^ 
5X  11=14/5^ 
5  X  12  =  5 


Table  of  Six-inch  Timber j  from  6  ^o  12  Inches  wide. 


Inches,        Feet. 
6X  6  =  3 
6X7  =  3i 
6X8  =  4 
6X9  =  4^ 


Inches,        Feet. 
6  X  10=:5 
6  X  11  =  5^ 
6X  12=:6 


EXERCISE. 

1.  "What  are  the  contents  of  a  piece  of  timber  5  inches  X 
5  inches  and  24  feet  long  ?  Ans.  50  feet. 

By  the  table  5X5  is  2|-  times  the  length,  for  the  con- 
tents ;  therefore  24  feet  X  2^^  =  2^  X  f  ii=W  =  50  feet 
in  board  measure. 

2.  Required  the  contents  of  a  joist  5X8  and  30  feet  long  ? 


30  feet  X  3^—100  feet. 


Ans.  100  feet. 


3.  Find  the  contents  of  a  beam  6  inches  X  8  inches  and 
36  feet  in  length  ?  Ans>  144  feet. 

36  feetX  4=  144  feet. 

4.  How  many  running  feet  of  6-inch  X  8-inch  timber  are 
equal  to  3,500  feet  running  lengths  of  5  X  12  inches  ? 

Ans.  4,375  feet. 
By  the  table  5  X  12  is  5  times  the  length,  for  the  con- 
tents, and  6X8  =  4  times  the  length  ;  therefore  3,500  feet 
X  5=  17,500  feet  =  contents  of  5  X  12  ;  then  17,500  -^ 
4  =  4,375  feet  =:  the  number  of  feet  in  length  of  6  X 
8  =z=  3,500  feet  of  5  X  12. 


ii 


ON  LUMBER   SURVEYING. 


17 


5.  What  will  abeam  cost  48  feet  long,  6  inches  by  11 
inches,  @  3^  cents  per  foot  ?  Ans.  $i).24. 

48  X  H  feet  =  264  feet  =  contents ;  264  X  3^  cents  = 
$9.24. 

Table  of  Timber  from  7  X  7  ^o  12  X  20. 


Seven-inch  Timber. 

Eight-inch  Timber. 

Nine-inch  Timber. 

Inches.  Feet. 

Inches.        Feet. 

Inches.        Feet. 

7X     7_4X 
7X8        4f 

8X8        5J- 

9X9        6f 

8X9        6 

9  X  10        7i 

7X    9  — 5i 

8  X  10  =  6f 

9X11  — 8i 

7  X  10        5^ 

8X11        7^ 

9  X  12  —  9 

7X  11  — 6i\ 

8  X  12  — 8 

7  X  12        7 

Ten-iuch  Timber. 

Eleven-inch  Timber. 

Twelve-inch  Timber. 

Inches.          Feet. 

Indies.        Feet. 

Inches.        Feet. 

10  X 10 —  ^ 

10X11—  4 

11  X  ii  =  10tV 

12  X  12  — 12 

11  X  12  —  11 

12  X  14  —  14 

10  X  12  — 10 

12  X  16 — 16 
12  X  18        18 
12  X  20=:  20 

1.  What  are  the  contents  of  a  piece  of  timber  12  by  12 
inches  and  30  feet  long?  Ans.  360  feet. 

2.  What  are  the  contents  of  a  beam  7  inches  by  9  inches 
and  30  feet  long?  Ans.  157|  feet. 

3.  Eec^uired  the  contents  of  a  piece  of  timber  9  X  10 
inches  and  40  feet  long?  Ans.  300  feet. 

By  the  table  9X10  =  7^  times  the  length  ;  40  feet  \  7^ 
=  300  feet. 

4.  In  2,500  feet  contents  of  9  X  10,  how  many  feet  run- 
ning lengths  of  9  X  10,  and  of  11  by  12  ? 

Ans.  Of  11  X  12,  227,3.  feet.     Of  9  X  10,  333^  feet. 

5.  What  is  the  cost  of  2,000  feet  running  lengths  of  12- 


ff, 


18 


SELF-INSTRUCTOR 


inch  by  20-inch  timber  @  3  cents  per  foot  of  board  measure  ? 
Ans.  $1,200.00. 

6.  Recjiuired  the  contents  of  a  piece  of  pine  timber  8  inches 
by  12  inches  and  24  feet  long?  Ans.  192  feet. 

7.  Wiuit  is  the  difference  in  feet  of  board  measure  be- 
tween 2,000  feet  running  lengths  of  9  X  12  and  2,000  feet 
running  lengths  of  12  X  12  ? 

Ans.  12  X  12  is  6,000  feet  more. 
By  the  table  12  X  12  =  12  times  the  length,  and  9  X 
12  =  9  times  ;  therefore  12  —  9  =  3  feet  difference  ;  2,000 
X  3  =  0,000  feet  difference. 

Example  showing  the   Manner  of  Drawing   or  Ruling   a 
Shingle  for  Plank  or  '2-inch,  also  the  Mode  of  Dotting. 

Rule.  —  Take  a  shingle  and  rule  it,  as  shingle  No.  1  is 
ruled,  the  dimensions  along  the  top  column,  and  the  lengths 
down  the  side  column ;  then  take  a  pencil  and  make  a  dot, 
thus  (,),  for  every  plank,  or  deal,  or  piece  of  timber,  as  the 
case  may  be.  Suppose  I  want  to  dot  a  2  X  6?  22  feet  long, 
3  times,  I  run  along  the  top  cohmm  of  dimensions  till  I 
come  to  2  X  G ;  I  then  go  down  said  line  till  I  come  oppo- 
site 22  in  the  column  of  lengths,  I  then  make  three  dots, 
thus  (,„).  Then  when  I  have  finished  dotting,  I  count  all 
the  dots,  and  place  the  figures  as  in  the  above  shingle  ;  those 
figures  I  afterwards  transfer  to  my  specification,  in  order  to 
find  the  contents  of  the  whole  quantity  of  pieces  I  have 
dotted. 

P.  S.  —  You  can,  if  reqiiired,  rule  your  shingle  so  as  to 
include  any  length  or  dimension,  and  most  shingles  are 
drawn  as  shingle  No  1  is. 


i  i 


[ 


measure  r 


ingles   are 


ON   LUMBER   SURVEYING. 
Plank  Shinglcy  No.  1. 


19 


. , 

Lengths.      =  § 

X 

X 

X 

X 

15 

X 

X 

•  • 

2 

9) 

X 

IM 

•  •• 

•  •• 

•  •• 

•  •• 

12 

•  ••• 

•  #•• 

8 

o 

X 

X 

CI 

2J 
X 

•••• 

4 

12 

18 

•  ••• 

4 

10 

5 

••••• 
•  ••• 

9 

6 

•  •••• 

6 

•  • 

2 

18 

•  ••t 

9 

5 

5 

•  • 

7 

•  ••• 

•  • 

6 

•  •• 

8 

• 

1 

14 

•  • 

2 

•  ••• 

4 
5 

•  ••• 

4 

• 

1 

2 

C 

••• 

C 

•  ••• 

9 

•• 

2 

16 

6 

•  • 

2 

5 

•  ••• 

4 

•  ••• 

4 

16 

6 

•  •• 

3 

10 

•  • 

2 

•  ••• 

•  ••• 

8 

•  • 

2 

•  ••• 

4 

10 

17 



6 

6 

10 

18 

■  ••• 

•  ••• 

12 

•  ••• 

4 

•    26 

• 

1 

•  ••• 

4 

•  •• 
3 

•  • 

2 

•  •• 

3 

5 

• 

1 

1 

5 

•  ••■« 

• 
6  ; 

i 

• 

•  •••    1 

1 

4 

•••• 

•  •• 

7 

19 

S 

5 

•  • 

2 

4 

• 

1 

20 

15 

•  •• 

3 

•  ••• 

4 

•  • 

2 

•  • 

2 

•  • 

•  • 

..  10 

• 

1 

21 

10 

•  • 

2 

• 

1 

•  • 

7 

•  ••• 

4 

•  • 

2 

• 

1 

•  • 

2 

22 

•  •• 

•  •• 

•  •• 

9 

•  ••• 

4 

•  ••• 

4 

•  •• 

3 

•  •• 

3 

•  •• 

3 

•  • 

2 

6 

■  —         - 1 

(^;ft 


!i 


20 


SELF-IXSTUUCTOR 


Example   of   Specification   of  the    Plank    Shinyle   No.   1, 
showing  the  manner  of  finding  the   Contents. 

Rule.  —  One  sixth  of  the  length  of  2-inch  stuff  multi- 
plied hy  the  width  will  give  the  contents  hi  feet  of  board 
measure  or  superficial  feet. 


Specification 

«/ 

Plank 

Shingle 

No.  1. 

Lengths.  S  § 

X 

18 
9 

X 

4 
5 

Mi 

X 

10 

5 

X 

15 
5 

X 

9 

7 

oo 

X 

2 
G 

X 

12 

8 

2 
X 

6 

8 

X 

IN 
6 

2 

X 

IN 

4 

1 

Coutcnta. 

12 

1,120 

13 

834 

14 

4 

4 

2 

5 

6 

9 

2 

623 

15 

2 

5 

1 

5 

2 

5 

4 

422 

16 

5 

3 

10 
6 

2 

_^ 

8 
6 

2     4 

|io 

4 

10 

1,000 

17 

482 

18 

12 

25 

3     2 

5 

5 

1,155 

19 

3 

4 

5 

2 

1 

4 

4     3 

1 

6 

792 

20 

15 

3 

4 

2 

1 

2 

16 

4 

1,180 

21 

10 
9 

2 

4 

1 
4 

7 
3 

4 

2 
3 

1 

3 

2 

1 
2 

7 
6 

885 

22 

828 

Total,  9,321  feet. 

Mule  for  calculating  a  2-Inch  or  Plank  Specif  cation. 

Multiply  the  number  of  pieces  or  dots  in  each  square  of 
the  table  by  the  width  of  said  pieces,  and  the  product  by  ^ 
of  the  length  for  the  contents. 


ON  LUMBER   SURVEYING. 


21 


To  find  the  Contents  of  Specification  Shingle,  No.  1. 

Multiply  the  number  of  pieces  in  each  square  of  the  table, 
opposite  the  first  lcn<,^th,  12  feet,  by  the  widths  of  the  differ- 
ent numbers  of  said  pieces,  and  then  by  ^  of  the  lenjj^th  for 
the  contents ;  thus,  for  the  first  column  running  parallel  to 
the  top  of  the  shingle, 


Breadth. 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

No.  Pieces. 

18 

4 

10 

15 

9 

2 

12 

6 

5 

4 

54     16    50     90     (53     16     108     GO     55     48 

Then  add  all  the  products,  54  -f  IG  +  50  -|-  00  -f  r.3  -|- 
16  -f  108  -I-  CO  4-  55  4-  48  =  560.  Then  1 2,  tiie  length, 
-H  6  ==  2  feet,  560  X  2  =  1,120  =  contents  of  the  first 
column.  Thus  proceed  until  the  contents  of  all  the  columns 
are  found,  then  add  the  whole  together  for  the  total  contents 
of  the  shingle. 

P.  S.  —  In  this  treatise,  when  there  is  a  fraction  of  half  a 
foot  over,  it  is  called  a  foot ;  when  less  than  half  a  foot, 
nothing. 

For  Joist  or  Scantling. 

Take  the  running  lengths  of  the  different  dimensions  and 
mark  down  every  100  feet,  then  add  np  your  shingle,  and 
multiply  the  different  sums  by  the  multiplier  of  each  dimen- 
sion, as  found  in  the  tables  for  the  contents  of  each.  Hem- 
lock joist  is  generally,  computed  by  this  plan. 


22 


SELF-INSTRUCTOR 
Joist  Shingle, 


2X8 

2^X3 

2X4 

2^X4 

2iX3 

2iX4 

8X4 

250 

100 

100 

10 

100 

100 

100 

100 

100 

90 

100 

100 

100 

250 

100 

100 

100 

100 

100 

100 

100 

50 

100 

100  . 

100 

100 

100 

100 

200 

100 

100 

100 

100 

100 

100 

100 

150 

100 

100 

100 

100 

100 

100 

100 

100 

100 

25 

100 

100 

100 

100 

150 

100 

100 

100 

200 

50 

100 

100 

100 

100 

613 

625 

667 

833 

450 

375 

900* 

*  The  numbers  at  the  foot  of  the  columns  are  feet  of  board  measure. 

3  inches  by  4  inches  by  the  table  is  once  the  length,  there- 
fore tliere  are  900  feet  of  3  X  4  contents.  There  are  in 
the  joist  shingle  500  feet  running  length  of  2^  X  4,  and 
2|  X  4  is  =  I  times  the  length ;  therefore,  500  -^  |= 
to  o75  feet  ==  contents  of  2^  X  4.  There  are  800  feet  run- 
ning lengths  of  2|  X  3,  and  2|^  X  3  is  ^^  times  the  length; 
therefoi'e,  800  -^  ^^j  =  450  =  contents.  There  are  1,000 
feet  of  2^  X  4;  therefore,  as  2^  X  4  is  §  of  the  length,  the 
contents  will  be  equal  to  1,000  -^  ^  =  833  feet.  Of  2 
inches  X  4  inches,  1,000  feet,  which  divided  by  |,  will  be 
the  contents  =  GG7  feet.  Of  2^  X  3  there  are  1,000  feet, 
and  2^  X  3  is  =  §  times  the  length  ;  therefore,  1,000  -j-  § 
=  625  feet.  Of  2  X  3  there  are  1,225  feet  running  lengths, 
and  2  X  3  is  ^  the  length;  therefore,  1,225  -i-  ^  =  612^ 
feet. 


ON   LUMBER   SURVEYING.  23 

New   Turk  Deal  Shingle,  S-Inch,  No.  2. 


Lengths,  s  § 

5* 

X 

X 

n 

X 

X 

O 

X 

09 

1^ 

X 

PS 

CM 

X 

14 

24 

14 

lA 



20 

IM 

20 

.'» 

15 

•  •••• 

5 

3 

1 
10 

•  ••• 

4 
7 

36 

20 

30 

11 
8 

IG 

12 

10 

17 

9 

15 

15 

8 

24 
5 

•  •• 

3 

•  ••• 

4 

27 

18 

••     2« 

8 

7 
7 

1) 
2 

•  •• 

3 

10 

19 

21 

12 

12 

1 

4 

20 

12 

4 

4 
4 

•  ••• 

4 

10 

21 

lU 

•  ••• 

4 

4 

5 

24 

4 

3 

7 

5 

27 
0 

22 

8 

•••••• 

12 

23 

15 

3 
8 

2r 

20 

4 

4 

30 

24 

4 

4 

4 

■  ■•• 

4 

25 

G 

3 

8 

10 

9 

26 

« 

4 

5 

0 
4 

5 

10 

10 

9 
5 

20 

•  ••• 

4 

27 

4 

10 

0 

28 

4 

5 

4 

4 

18 

29 

5 
4 

11 

9 

12 
5 

10 

30 
24 

30 

8 

21 

9 

2ft 

P.  S.  —  New  York  deiil  in  from  12 feet  up  in  length,  and  from  6  to  12  inuhes  wide,  and  must 
be  good  aprucc  lumber,  free  from  cracks,  rots,  or  large  knotx,  etc. 


t 


:  i 


jI  :I 


24  SELF-INSTRUCTOR 

Specification  of  New  York  Deal  Shingle,  No.  2. 


0  .; 

«o 

t- 

CO 

Ci 

o 
1-^ 

Lengths.     5  g 

X 

X 

X 

X 

X 

X 

X 

Contents. 

■p  * 

24 

eo 
14 

P5 

20 

eo 
16 

e»s 
18 

eo 
20 

eo 
30 

14 

4,571 

15 

36 

20 

30 

11 

5 

3 

4 

3,098 

16 

12 

6 

8 

10 

1 

7 

1,548 

17 

9 

24 

15 

15 

16 

27 

4,420 

18 

26 

8 

5 

9 

8 

7 

10 

2.744 

19 

21 

12 

3 

2 

1 

7 

4 

1,838 

20 

12 

12 

4 

3 

4 

4 

10 

2,095 

21 

16 

27 

4 

4 

4 

5 

2,667 

22 

8 

4 

3 

5 

4 

3 

12 

1,991 

23 

27 

20 

24 

3 

7 

20 

5,089 

24 

15 

4 

4 

8 

4 

4 

2,070 

25 

6 

10 

4 

4 

3 

9 

8 

2.494 

26 

5 

10 

10 

9 

20 

4 

6 

3,750 

27 

10 

9 

4 

5 

4 

5 

4 

2,315 

28 

18 

4 

5 

4 

4 

5 

2,429 

29 

30 

11 

9 

12 

5 

10 

4,40  i 

30 

24 

20 

8 

5 

4 

21 

9 

5,790 

Rule  for  finding  the  Contents  of  S-Inch  Deals. 

Multiply  ^  of  the  length  of  the  deals  by  the  breadth  of 
them,  for  the  contents. 

This  shingle  is  done  the  same  way  as  the  plank  shingle 
No.  1,  excepting  that  ^  cf  the  lengths  are  taken  instead  of  ^ 
of  them. 


ON  LUMBER   SURVEYING. 
New  York  Deal  Shingle,  A-Jnck,  No.  3. 


25 


1 

CO 

t» 

00 

CJ 

o 

1— 1 

Lengths. 

X 

X 

X 

X 

X 

X 

X 

•* 

t 

^ 

"^ 

■^ 

■<t 

■^ 

14 

1  ,,,,,,,,, 

32 

10 

•••     30 

•  • 

12 

14 

17 

5 

15 

• 

• 

24 

10 

11 

iO 

16 

21 

16 

•  •«• 

••••••• 

•  ••• 

•  t«« 

4 

7 

4 

5 

6 

4 

13 

17 

^ 
* 

••     11 

Iri 

14 

24 

•  ••• 

9 

18 

•  • 

•  «•• 

•  ■• 

7 

2 

1 

4 

3 

5 

4 

19 

7 

4 

•  ••• 

4 

•  ••• 

4 

IG 

5 

20 

8 

6 

•  •• 

3 

•  • 

2 

10 

12 

21 

•  •«• 

•  «•• 

•       25 

8 

4 

6 

5 

4 

22 

6 

7 

7 

9 

12 

11 

23 

8 

•  •• 

3 

•• 

2 

•  •• 

3 

• 

1 

•  • 

2 

•  •• 

3 

•  ••• 

•  •• 

•  •• 

24 

6 

4 

6 

' 

3 

••       11 

6 

JRule  for  Jinding  the  Contents  of  \-lnch  Deals. 

Multiply  the  length  divided  by  3  by  the  breadth  for  the 
contents  iu  feet  of  board  measure. 


H  '-^ 


26 


SELF-INSTRUCTOR 


What  are  the  contents  of  32  pieces  14  feet  long  and  4  X  G  ? 
32  X  14  =  448  feet  of  running  length,  then  14  -j-  3  =  4§ 
=  ^  of  length  of  each  piece.  And  4  X  C  inches  by  the 
table  is  =  2  times  the  length,  for  the  contents,  therefore 
448  X  2  =  81)0  feet  =  contents.  By  taking  ^  of  the 
length,  it  is  done  thus,  32  pieces  X  6,  their  breadth  = 
l'J2  X  4§  =  896  feet,  contents.  Or  multiply  the  number 
of  ])ieces  b}^  the  length  of  one,  and  the  product  by  g-  of  the 
width  of  the  deals  tor  the  contents  of  4-iuch. 


Specif  cation  of  New  York  Deal  Shingle,  4-Inck,  No.  3. 


1 

0  ■/■ 

» 

t- 

QO 

«i 

0 

fH 

Lengths.     £  5 

X 

X 

X 

X 

X 

X 

X 

Contents. 

0  "■ 

32 

10 

30 

12 

14 

17 

5 

14 

4.653 

15 

24 

10 

11 

10 

16 

21 

4,150 

16 

4 

7 

4 

5 

5 

4     13 

2,133 

17 

9 

7 

11 

18 

14    24 

4,709 

18 

7 

2 

1 

4 

3 

5 

4 

1,398 

19 

7 

4 

4 

4 

10 

5 

1,887 

20 

8 

6 

3 

2 

10 

12 

2,607 

31 

25 

8 

4 

6 

5 

4 

2,891 

22 

6 

7 

7 

9 

12  ;  11 

3,777 

23 

8 

3 

2 

3 

1 

2 

3 

1,380 

24 

6 

4 

5 

3 

3 

11 

6 

2,832 

Total,  32,417  feet.  | 

ON  LUMBER   SURVEYING. 


27 


Solution  of  Specification  No.  3. 


No.         Br.         ] 

Products. 

No.         Br. 

'roducts. 

No.         Br.        ] 

Products. 

32  X     6 

192 

9X     G 

54 

8X6 

48 

10  X     7 

70 

8  X     7 

56 

6  X     7 

42 

5  X   12 

60 

11  X     9 

99 

3  X     9 

27 

30  X     8 

240 

18  X  10  = 

180 

2  X  10 

20 

12  X     9 

108 

14  X  11 

154 

10  X  11 

110 

14  X   10 

140 

24  X   12 

288 

12  X   12 

144 

17  X   11  — 

187 

831 

391 

997 

17  -^    3 

5§ 

20-7-3 

6| 

^A    •      ^  — 

4^ 

^3 

Contents, 

4,709 

Contents, 

2,607 

Contents, 

4,653 

7  X     6 

42 

25  X     6  — 

150 

24  X     6 

144 

2  X     7 

14 

8X8 

64 

10  X     7 

70 

1   X     8 

8 

4X9  — 

36 

11  X     8 

88 

4X9 

36 

6  X   10 

60 

10  X  10 

100 

3  X   10 

30 

5  X    11  — 

55 

16  X  11 

176 

5  X   11 

55 

4  X  12 

48 

21   X  12 

252 

4  X  12 

48 

830 

233 

21  H-    3 

413 

7 

15     '       3  — 

5 

18  H-    3 

6 

Contents, 

2,891 

Contents, 

4,150 

Contents, 

1,398 



_ 



6X7  — 

42 

4  X     6 

24 

7  X     6 

42 

7  X     8 

56 

7  X     7 

49 

4  X     7 

28 

7X9  — 

63 

4X8 

32 

4  X     8 

32 

9  X   10  — 

90 

5  X     9 

45 

4X9  — 

36 

12  X  11 

132 

5  X   10 

50 

10  X  10 

100 

11   X   12— 

1.32 

4  X   11 
12  X  13  — 

44 

5  X  12 

60 

15fi 

515 

298 

22  -T-     3 

400 

19  -^     3  -t:: 

^h 

1«  -.•.     3  — 

5I 

Cnti  fonts 

3,777 

"3 

1,887 

Contents, 

2.133 

I' 


'fr 


28 


SELF-INSTRUCTOR 


Solution 

of  Specification 

No.  3.  - 

-  (  Continued.) 

No.         Br.         Products. 
8X6               48 
3  X     7  —        21 
2X8  —        16 
3X9              27 

1  X   10              10 

2  X   11              22 

3  X  12              36 

No.         Br. 
6X6: 
4X7: 

5  X     8: 
3  X     9: 
3  X   10: 

11   X    11: 

6  X   12: 

24  -^     3: 

Contents, 

Products. 

=       36 
=       28 
=       40 
=       27 
=       30 
121 
=       72 

354 

—         8 

24  feet  being  the 
length  of  the  pieces 
in  the  hist  column, 
I  tak(?  the  ^  of  it 
8,   and  multiply  it 
by   the  product  of 

23-7-3=: 

180 
7f 

the   No.    of    pieces 
and  their  breadths. 

Contents,       i,380 

2,832 

Rule  for  computing  5-inch  Timber. 

Multiply  the  number  of  pieces  in  each  square  of  the 
shingle,  by  their  width  as  given  in  the  top  column,  and  the 
product  by  the  length  divided  by  2^  for  the  contents. 

By  multiplying  the  length  of  a  5-inch  stick  by  the  width 
of  the  same,  and  the  product  by  the  length  divided  by  2g, 
you  will  get  the  contents  in  feet  of  Board  Measure. 

Required  the  contents  of  33  pieces  10  feet  long  of  5 
X5. 

1st  Solution.  —  33  X  1^  =  330  X  2yV  =  687^  feet. 

2c?  Solution.  —  Find  the  contents  of  10  pieces  33  feet 
long  and  5  by  5.  10  X  5  =  50,  33  -h  2§  =  ^\  X  ¥  = 
•t^  =13f,  therefore  50  X  13f  =  687^  =  Ans. 


ON  LUMBER  SURVEYING. 


29 


Timber 

Shingle  Five-inch,  No.  A 

• 

Dimen- 
sions. 

X 

X 

X 

X 

W5 

X 

in 

o 

X 

Its 

X 

in 

i-t 

X 

in 

20 

7 

2 

5 

16 

2 

• 

1 

6 

•  •••• 

6 

21 

8 

•••••• 

6 

5 

8 

•  •••• 

5 

6 

•••• 

4 

•  ••• 

4 

22 

6 

5 

5 

■  • 

2 

5 

•  •• 

3 

5 

5 

•• 

2 

5 

23 

6 

•  • 

2 

5 

8 

8 

24 

•—  20 

7 

•  •• 

3 

8 

•  •• 

3 

•  •• 

3 

15 

9 

25 

«• 

2 

• 

1 

• 

1 

2 

•  ••• 

4 

4 

5 

26 

7 

3 

2 

6 

1 

•  ••• 

4 

•  ••• 

4 

•  • 

2 

• 

1 

30 

....  j2 

3 

•  •• 

3 

••     10 

•  •• 

3 

•  • 

2 

31 

7 

5 

•  •• 

3 

•  •• 

3 

•  •• 

3 

••• 

3 

6 

9 

32 

7 

•  •• 

3 

•  • 

2 

•  ••• 

4 

•  •• 

3 

•  •• 

3 

•  • 

2 

* 

1 

33 

••   10 

...» 
4 

•  • 

2 

• 

1 

3 

5 

9 

'ff^^ 


mmmmm 


m  \ 


30 


SELF-INSTRUCTOR 


Specification  of  Five-inch   Timber  Shingle,  No.  4. 


1 

C3  00 

\a 

» 

»«• 

flO 

o> 

o 

1^ 

«»j 

Lengths.       |  g 

X 

X 

X 

X 

X 

X 

X 

X 

Contents. 

5  '■ 

Ift 

in 

lA 

to 

«o 

in 

lA 

in 

20 

7 

2 

5 

16 

2 

1 

6 

5 

2,941 

21 

8 

6 

6 

8 

5 

5 

4 

4 

3,167 

22 

6 

5 

6 

2 

3 

6 

5 

5 

2,778 

23 

6 

2 

5 

5 

8 

2 

8 

3,191 

24 

20 

7 

3 

8 

3 

3 

15 

9 

6,570 

25 

2 

1 

1 

2 

4 

4 

5 

1,865 

26 

7 

3 

2 

1 

4 

4 

2 

1 

2,004 

30 

12 

6 

3 

3 

10 

3 

2 

4,025 

31 

7 

6 

3 

3 

3 

3 

6 

9 

4,405 

32 

7 

3 

2 

4 

3 

2 

1 

2,387 

33 

10 

4 

2 

1 

3 

3 

5 

9 

4,345 

Total,  36.678  feet. 

Example,  shoioing  how  to  compute  a  d-inch  Specijication. 


No.       Br. 

No.       Br. 

No.        Br. 

No.        Br. 

7  X    5  —    35 

8  X     5  —  40 

6X5        30 

6  X    6  —  36 

2  X    6  —    12 

6  X    6  —  36 

5X6        30 

2X7        14 

5  X    7  —    35 

5X    7  —  35 

5X7        35 

5  X    8  —  40 

8  X  16        128 

8X8        64 

2X8        16 

5X9        45 

2X9          18 

5X    9—45 

3X9        27 

8  X  10        80 

1  X  10 —    10 

5  X  10        50 

5  X  10        50 

2X11        22 

5X11          55 

4  X  11        44 

5  X  11  —  55 

8  X  12        96 

5  X  12          60 

4  X  12 — 48 

5  X  12        60 

353 

362 

303 

333 

*20-=-2*  — 8^ 

21-^2|        8| 

22-^2^  — 9^ 

23-h2^_9TV 

Contents,  2,941 

Contents,  3,167 

Contents,  2,778 

Contents,  3,191 

*  20  feet,  the  length  of  tho  pieces,  divided  by  2g,  and  the  result,  8},  multiplied 
by  353  =  2,941  feet  =  contents  of  20  ft  pieces. 

Invert  :i^  =  ^^XY  =  ¥lI^  =  8i 


ON   LUMBER   SURVEYING. 
Timber  Shingle,  Six-inch,  No.  5. 


81 


1 

Lengths,    gg 

g-E 

CO 

X 

CO 

X 

CO 

00 

X 

CO 

X 

CO 

10 

o 

X 

CO 

II  X  9 

X 

CO 

20 

•  •• 

8 

5 
5 

10 

10 

...       14 

18 

3 

21 

7 

7 

•  ••• 

4 

1 

9 
11 

....      20 
....      15 

22 

7 

6 

6 

•  ••#• 

5 

23 

7 

& 

•  •• 

3 

6 

11 

22 

....      15 

24 

•  •• 

8 

4 

•  ••• 

4 

•  •• 

8 

8 

5 

25 

6 

6 

•  ••• 

4 

t 

9 

10,     "     2o! 

1 

26 

7 

•  •• 

8 

4 

5 

4 

11 

1 

...       14 

27 

•  •• 

8 

3 

6 

5 

6 

10 

14 

28 

10 

8 

•  ••• 

4 

4 

& 

G 

6 

29 

7 

6 

4 



•  •• 

3 

6 

9 

30 

7 

4 

•  • 

2 

6 

5 

5 

6 

L- —                             .                                                                          1 

Ride  for  finding  the  Contents  of  Q-inch   Timber. 
Multiply  the  number  of  pieces  or  dots  by  the  width  of 
said  pieces,  and  tlieu  multiply  the  product  by  half  the  length 
of  one  of  the  pieces,  for  the  contents. 


^in 


¥ 


82 


SELF-INSTRUCTOR 


\m 


j 


What  are  tho  contents  of  18  pieces  of  6  X  6,  and  20  feet 
long?  18X  C:=108;  20 -^  2=  10, 108  X  10=  1,080  feet. 
By  the  Tabic  6  X  6  is  three  times  the  length  for  the  con- 
tents, therefore  20  X  18  =  3G0  feet  runnhig  length;  360  feet 
X  3  feet  =  1,080.  Ans.  1,080. 

So  we  find  the  same  result  by  both  rules. 

Specification  of  Timber  Shinrile,  No.  5. 


,      1         1         1            O   1  ^ 

(N 

a  « 

o 

t-- 

CO 

0) 

>-• 

.-( 

fH 

Lengths.       g  © 

X 

X 

X 

X 

X 

X 

X 

Contents. 

(5" 

CO 

18 

(O 

3 

3 

10 

CO 

5 

CO 

10 

CO 

14 

20 

5.710 

21 

7 

7 

4 

5 

9 

26 

6,321 

22 

7 

6 

6 

5 

10 

11 

15 

6.358 

23 

7 

5 

3 

6 

11 

22 

15 

7,900 

24 

3 

4 

4 

3 

3 

6 

2,544 

25 

9 

6 

6 

4 

10 

20 

6,712 

26 

7 

3 

4 

5 

4 

11 

14 

6.097 

27 

3 

3 

5 

5 

6 

10 

14 

6,237 

28 

10 

3 

4 

4 

5 

6 

6 

4,718 

29 

7 

6 

4 

5 

9 

3 

5 

4,988 

30 

7 

4 

2 

6 

5 

5 

6 

4,755 

Total,  62,340  feet. 

Examples  showing  how  to  compute  the  Specification  No.  5  of 

6-inch  Timber. 


Br.        No. 

Rr.        No. 

Br.       No. 

Br.       No. 

6  X  18 — 108 

7  X    7  —    49 

6X     7r 

—    42    7  X    6 —    42 

7X3          21 

8X7          56 

7X6  = 

—    42    7  X    5  —    35 

8X3          24 

9  X    4  —    36 

8X    6r 

48    8  X    3          24 

9  X  10          90 

10  X    5  —    50 

9X    5z 

—    45    9  X    6 —    54 

10  X    5  —    50 

11  X    9 —    99 

10  X  10  = 

100  10  X  11        110 

11  X  10  —  110 

12  X  26       312 

11  Xii  = 

—  121  11  X  22  —  242 

12  X  14        168 

12  X  15  = 

=  180  12  X  15  =  180 

571 

602 

578                       687 

20  -r-  2          10 

21-^2 —   10^ 

22 -i- 2  = 

Contents, 

11    23-^2          11^ 

Contents,  5,710 

Contents,    6,321 

6,358    Contents,   7,900 

ON  LUMBER   SURVEYING. 


88 


What  is  the  cost  of  a  piece  of  pine  timber  6  inches  X  10 
inches,  and  38  feet  in  length  @  31  cts.  per  foot  ? 

^  ,    .  Jns.  $6.65. 

Solution, ^Length  38 --2  =  10;  19  X  by  the  breadth 
10  =  190  feet,  contents.     190  feet  @  3h  =  $6.65. 

By  the  Second  liule.     6  inches  X  i"o  inches  =  5  times 
the  length,  for  the  contents,  therefore  38X5  =  190  feet 
190  feet  X  3^  cts.  =  $6.65. 

Hitk  for  finding  the  Contents  of  l-inch    Timber. 
Multiply  the  width  by  the  length,  divided  by  If 

Required  the  contents  of  a  piece  of  timber  7X7  and  20 
feet  lonof  ? 

Divide  the  length,  20  feet,  by  1^20  -M^  =  il|),  and 
multiply  the  breadth,  7  inches,  by  the  quotient,  11§. 

!•'!=¥;  ¥Xl  =  ^$i  =  81f  feet  =  content8  in  su- 
perficial feet. 

2d  Operation.  -  By  the  table  7  X  7  is  =  to  4tV  times 
the  length,  for  the  contents,  therefore  20  feet  X  4tV  =  81^ 
feet  =  contents.  ^ 

Timber  is  oflen  surveyed  and  the  contents  marked  on 
each  piece,  and  then  put  down  on  a  shingle  for  contents  in 
Its  proper  column. 


0. 

6  — 

42 

5  = 

35 

3  — 

24 

6  = 

54 

1  — 

110 

9. — 

242 

5 

180 

3 


84 


SELF-INSTRUCTOR 


Timber  Shingle 

,  Scven-inch. 

iVb.  6,  and 

Speeiji 

cation. 

Lengths,  g  3 
5* 

I* 

X 

24 

00 

X 
1^ 

9> 

X 

o 

X 

X 

X 

ContentH. 

20 

IG 
6 

•  ••• 

4 

7 
8 

•  ••• 

4 

5 

16 

6 

9,321 

18 
8 
8 

21 

7 

5,059 

9 

22 

4 

6 
7 
6 
6 
6 

8 

10,062 

23 

.... 

4 

7 

7 
8 
5 

•  ••• 

4 

•  ••• 

4 

10,062 

9 
21 

18 

27 

24 

8 

•  ••• 

4 

7,420 

9 

25 

•  ••• 

4 

•• 

2 

6 

3,807 

26 

14 

6 

6 

7 

1,493 

27 

5 

8 



15 

8 

6 
4 

5 

7 

5,197 

28 

7 

•  ••• 

4 

8 

8 

4 

5,390 

29 

7 

7 

6 

•  • 

2 

5,988 

30 

4 

5 

7 

8 
3 

6,755 

31 

10 

12 

•  ••• 

4 

•  • 

2 

• 

1 

5,624 

i;  ! 


ition. 

^ontentH. 

9,321 

5,059 
10,062 
10,062 
7,420 
3,807 
1,493 
5,197 
5,390 
5,988 
6,755 
5,624 


ON  LUMBER  SURVEYING. 
Timber  Shingle^  Eiyht-inch.     No.  7. 


85 


1 

1 

g  x 

Lengths.          g  5 

5* 

00 

X 

OO 

9> 

X 

CO 

o 

FN 

X 

QO 

II  X8 

2 

X 

26 

7 

12 

•  •••• 

6 

12 

9 

18 

27 

5 

8 

•  •• 

3 

8 

•  ••• 

4 

28 

•  • 

2 

•  •• 

3 

6 

5 

29 

•  ■••• 

5 

•  ••• 

4 

•  •• 

8 

•  • 

2 

•  •••• 

6 

30 

••• 

3 

8 
8 
4 
7 

10 

10 

81 

6 

• 

1 
4 

•  • 

2 

•  • 

2 

32 

•  ••• 

4 

•  • 

2 
5 

•  •• 

3 

7 

33 

5 

•  •• 

8 

34 

•  « 

2 

6 

•  •• 

3 

•  •••• 

5 

•• 
2 

S5 

5 

9 

7 

36 

6 

12 

9 

37 

7 

15 

.S2 

24 

21 

Rule  for  jinding  the  Contents  of  S  bg  S   Timber. 
Divide  the  length  by  1^,  and  multiply  the  quotient  by  the 
width  of  the   timber   for   the   contents   in   feet   of   board 


measure. 


'ff-— ■ 


86 


SELF-INSTRUCTOR 


■li 


Example  showing  how  the  firt  column  of  8-hich  specifi- 
cation is  (K)iic. 


26-f-l^=-). 
Invert   tlic  divisor, 


Br. 

Nu.  plecefl  each  20  feet  long 

8X 

12 

—    DO 

9X 

IH 

—  1G2 

10  X  12: 

—  120 

11  X 

9 

—    1)9 

12  X 

7 

—    84 
5G1 

2G  : 

H-- 

=    14 

2244 
5G1 

7854  feet=  contents. 
Specification  Shinyle,  Eight-inch.     No.  7. 


g  ^ 

00 

05 

o 

1-^ 

Lengths,    g  g 

X 

X 

X 

X 

X 

Contents. 

(5  ■■' 

oo 

00 

18 

GO 

QO 

00 

26 

12 

12 

9 

7 

7,854 

27 

5 

5 

3 

3 

8 

4,392 

28 

2 

3 

6 

5 

4 

3.845 

29 

5 

4 

3 

7 

5 

4,698 

80 

10 

3 

2 

10 

8 

6,660 

81 

5 

2 

2 

8 

3,782 

82 

4 

1 

2 

3 

4 

3,029 

33 

5 

4 

5 

7 

7 

6,314 

84 

3 

2 

6 

5 

2 

4,103 

85 

5 

3 

9 

4,060 

86 

12 

6 

7 

9 

8,040 

37 

15 

7 

24 

21 

32 

38,406 

Total,  9,6183  feet. 

m 


l83 


feet. 


ON  LUMBKR  SURVEYING.  87 

Timber  Shingle,  Nine-inch.     No,  8. 


til 
LtiiigtiiH.        a  5 

g-2 

!    6X6 

i 

o 
X 

9) 

II  X  6 

IN 

X 

5 

26 

6 

6 

u 

27 

la 

6 

6 

18 

28 

•  • 

2 

4 

••• 

3 

•  ••• 

4 

«• 

29 

8 

•  • 

2 

•  ••• 

4 

30 

6 

•  • 

2 

•  ••■ 

4 

5 

31 

8 

•  •• 

8 

4 

4 

32 

5 

•• 

2 

•• 

2 

15 
3 
3 
6 

•  a* 

3 

33 

• 

1 

6 

34 

•• 

2 

•  • 

2 

4 
2 

•  ••* 

4 

•  •••« 

6 

• 

1 

35 

2 

36 

6 

Hide  for  Jinding  the  Contents  of  Nine-inch   Timber. 

Divide  the  length  by  1^  and  multiply  the  quotient  by  the 
breadth  of  the  stick  for  the  contents. 

Required  the  contents  of  a  piece  of  timber  9  X  12  inches 
and  26  feet  long? 

26 -J- ^=  10^.     191  X  12  =  234  =  contents. 


li 


il;|l 


6S 


SELF-INSTRUCTOR 
Timber  Shingle,  Teii-inch.     No.  9. 


• 

Lengths.             g  § 

10  X  10 

10  X  11 

10  X  12 

26 

5 

13 
5 

36 

27 

•  ••• 

4 

4 

28 

6 

•  •• 

3 
6 

•  • 

2 

•  •• 

8 

11 

11 

•  ••• 

4 

29 

8 

30 

•  ••• 

4 
6 

4 

• 

1 

31 

32 

••• 

8 

6 

27 
6 

••• 

3 

• 

1 

33 

5 

•• 

2 

•  • 

2 

34 

6 

35 

5 

36 

12 

25 

2\ 

1 

12 

5 

4 

11 

• 

4 

•  • 

4 

• 

1 

•  • 

3 

>•• 

5 

•  ••• 

6 

■'i 


ON   LUMBER   SURVEYING.  39 

Specification  of  Timber  Shingle,  Nine-inch.     No.  8. 


1                                                                                                                                                                                                                             1 

pr 

CO 

Dimen- 
sions. 

X 

o 

X 

0) 

1-* 

X 

(N 

X 

C5 

Contents. 

26 

6 

14 

6 

5 

6.240 

27 

18 

12 

5 

5 

8,039 

28 

2 

3 

4 

2 

2,436 

29 

4 

3 

2 

4 

2.958 

30 

6 

2 

5 

3,195 

31 

8 

3 

4 

4 

4,510 

32 

5 

4 

2 

15 

6,888 

33 

2 

1 

5 

3 

2,945 

34 

2 

2 

4 

3 

3,009 

35 

2 

4 

5 

5 

4,541 

36 

6 

2 

1 

3 

3,267 

Contents,  48,028  feet. 

1 

pi 


Specification  of 

Timber 

Shingle,   Ten4nch.     No.  9. 

1 

Lengths.    |  § 

5* 

o 

X 

o 
36 

FN 

X 

© 

I-H 

X 

o 
5 

Contentj. 

26 

13 

12,198 

27 

4 

5 

4 

3.297 

28 

11 

5 

11 

6.9:J0 

29 

8 

3 

4 

3,891 

30 

4 

6 

4 

3,850 

31 

6 

2 

1 

2,428 

32 

27 

3 

3 

9.040 

33 

5 

5 

5 

4,587 

34 

3 

2 

6 

3,513             , 

35 

1 

2 

5 

2,683 

36 

12 

25 

24 

20,490 

Contents,  72,907  feet. 

*  '^isa 


{  IHllll 

t        I     IMI  I  'I' 


40 


SELF-INSTRUCTOR 


Hide  for  Ten-inch   Timber. 

Divide  the  length  by  1^  and  multiply  the  quotient  by  the 
breadth,  for  the  contents  in  feet  of  board  measure. 

Reciuired  the  contents  of  a  stick  36  feet  long  10  inches  by 
11  inches? 

3G-J-  U==30,  and  30  X  11  =  330  feet  =  contents. 

2d  Solution.  — By  the  table  10  X  H  is  9 J  times  the 
length,  for  the  contents;  therefore,  36  feetX  ^^h  =  330  feet 
=  contents. 

Examples  showing  how  9  and  10  inch  specifications  are 
made  out. 

Nine^inch. 

Br.       Pieces.    Pro. 

9  X    6=    54 
10X14  =  140 

11  X    6=    66 

12  X    5=    60 


S20 


Ten-inch 

• 

Br. 

Pieces. 

Pro. 

10X36  = 

360 

11 

X13  = 

143 

12 

563 

X    5- 

GO 
563 

2 

21 

26-^  U  = 


191 


2880 
320 
160 

Contents  =  6240 
Length,  26  -^  1^;  1^  = 
§.     Inverted  : 


3)1126 


375 


563 
1126 
375 


|;|X  ¥ 


12,198  feet 
Length,  26  -f- 1 1 ;   1  ^  = 
^  =  inverted  to  g  ;  ^  X  ¥ 

130  91  2 


7rt  lOi 

4    —  ^"^a* 


P.  S.  —  All  the  specifications  in  this  book  are  done  in  a 
manner  similar  to  the  specification  of  the  Plauk  Shingle 
No.  1. 


ON  LUMBER   SURVEYING. 
Eleven-inch  Shingle  No.  10. 


41 


Lengths. 

1 

E.2 

pH 

X 

IN 

X 

20 

24 

30 

21 

6 

4 

22 

3 

9 

23 

4 

•  ■ 

2 

24 

•  •••• 

• 

1 

■ 

6 

•  •• 

3 

26 

• 

1 

•  • 

2 

27 

5 

1 

28 

6 

•• 

2 

29 

6 

4 

30 

6 

I 

a 
o 

pC 


^ 


i.       :i 


-§ 


<4i 


a» 


^ 

« 

'_^\ 

s 

>-. 

•Ka 

j^ 

■4«* 

O 

•^ 

03 

"a 

X 


a 


^ 


O     CO 


^  o 

•ill 

=  x 


C;    GO 


■*->    o    C    " 

^  .s  1  -- 


■fT 


•i'S' 


(iii 


!|l 


P 


III!  I  III 


lilllli 


tiii^Ml' 


!  flP'f 


42 


SELF-INSTRUCTOR 
Timber  Shingle,   Twelve-inch,  No.  11. 


1 

Lengths.          g  § 

5" 

X 

l-H 

X 

l-H 

l-H 

X 

CO 

l-H 

X 

o 

X 

20 

•••• 

4 

•••• 

4 

25 

16 

ic 

21 

•  •• 

3 

•  • 

2 

• 

1 

• 

1 

•• 

2 

22 

4 

• 

1 

•  • 

2 

•  • 

2 

••• 

8 

23 

•  • 

2 

• 

1 

•  • 

2 

• 

1 

24 

8 

•  •• 

3 

8 

1 

2 

25 

•  ••• 

4 

■ 

1 

• 

1 

•  •• 

3 

• 

1 

26 

• 

1 

2 

•  • 

2 

• 

1 

•••• 

4 

27 

•  • 

2 

•  •• 

3 

•  •• 

3 

•  ■ 

2 

••• 

8 

28 

• 

1 

•  • 

2 

•  •• 

3 

4 

4 

29 

•  ••• 

4 

•  •• 

3 

• 

1 

• 

1 

•• 

2 

80 

•  • 

2 

2 

•  •• 

3 

•• 

2 

5 

.:j 

Jiule  for  Ttvelve-inch   Timber. 

Multiply  the  length  hy  the  width  for  the  contents  in  feet. 
Required,  the  contents  of  16  pieces  of  12  X  -^^  hich  timher, 
and  20  feet  long?  16  X  20=320.  320  X  20  =  6,400 
feet  =  contents  in  feet  of  board  ineasui-e. 


ON  LUMBER   SURVEYING. 
Specification  of  Shingle  No.  10. 


43 


o 

X 

16 

•  « 

2 

•  •t 

8 

• 

1 

•  • 

2 

• 

1 

•  ••• 

4 

•  •• 

3 

•  ••• 

4 

•  • 

2 

•  •••• 

1 

Lengths,  g  3 

g-3 

11  X  11 

11  X  12 

Contents. 

20 

24 

36 

12.760 

21 

6 

4 

2,194 

22 

9 

3 

2.722 

23 

4 

2 

1.434 

24 

5 

1 

1.774 

25 

5 

3 

2,085 

26 

1 

2 

834 

27 

5 

3 

2,252 

28 

5 

2 

2.028 

29 

6 

4 

3.030 

30 

5 

2 

2,272 

Total,  33.285 

Specific 

?a<io?i  of  Shingle  No. 

11. 

1 

Lengths,  g  % 

g'2 

IN 

X 

(N 

I— 1 

X 

X 

00 

X 

(N 

o 

X 

Contents. 

20 

25 

16 

4 

4 

16 

19,600 

21 

3 

2 

1 

1 

2 

2,898 

22 

4 

1 

2 

2 

3 

4.180 

23 

2 

1 

2 

1 

2,162 

24 

8 

3 

8 

1 

2 

7,776 

25 

4 

1 

1 

3 

1 

3.800 

26 

1 

2 

2 

1 

4 

4,420 

27 

2 

3 

3 

2 

3 

5  670 

28 

1 

2 

3 

4 

4 

6,720 

29 

4 

3 

1 

1 

2 

4,756 

30 

2 

2 

3 

2 

5 

7,080 

Total 

,  69,071 

yoa 


44 


SELF-INSTRUCTOR 


I 


hi 


Hide  for  finding  the  Contents  of  Battens  or  TtvO'Cnd-a- 

Half-inch   Stuff. 


Inch. 

Inch. 

What  are  the  contents 

2i  X  2  _    A 

2iX    8  = 

=  1^ 

of  a  batten  22  ft.  long  2^ 

2^  X  3  -^     f 

2iX    9  = 

=  1& 

niches  by  12  inches  ? 

2iX  4            f 

2^X  10  = 

=  2,V 

By  this  rule  2^  X  12 

2iX5_l5>4 

2^X11  = 

=  2,7^ 

is —  2^  times  the  lenjrth, 

2^  X  6        \\ 

2iX  12  = 

=  21 

for  the  contents,  therefore 

2iX7_lJi 

22  ft.  X  '^h  —  »'5ft-  ^ns. 

Batten  Shingle,  No.  12. 


OR 

Dimen- 
sions. 

X 

IN 

t* 

X 

<N 

X 

Him 

IN 

X 

IN 

o 

X 

<N 

X 

IN 

(N 

PH 

X 

IN 

20 

15 

8 

•  ••• 

4 

•  • 

2 

12 

4 

•  ••• 

4 

....    45 

21 

8 

••• 

3 

•  ••• 

4 

•  ••• 

4 

•  •• 

3 

8 

22 

4 

•  •• 

8 

•  •• 

8 

•  • 

2 

• 

1 

••• 

3 

9 

23 

•  ••• 

4 

3 

•••• 

4 

4 

•  •• 

3 

•  •• 

3 

4 

24 

12 

8 

•  ••• 

4 

•  • 

2 

• 

1 

•  ••• 

4 

12 

25 

6 

•  ••• 

4 

••• 

8 

8 

••• 

3 

•• 

2 

9 

26 

8 

•  ••• 

4 

•  •• 

3 

•• 

2 

• 

1 

•  •• 

3 

....     24 

ON   LUMBER  SURVEYING. 


45 


(M 


X 


12 


Rule  for  Jinding  the  Contents  of  Battens. 

Divide  the  length  of  the  piece  by  4^,  and  multiply  the 
product  by  the  breadth  of  the  i)iece,  for  the  contents  in  feet ; 
or  multiply  the  length  by  the  number  given  in  the  table  for 
the  contents.  Ans.  80  feet. 

What  aie  the  contents  of  a  batten  24  feet  long  2^  by  G  ? 

2^  X  G,  by  the  Table,  is  =  to  1  ^  times  the  length  ;  24  X 

=  80  feet. 

Second  Solution.  —  24  -j-  4^  =  5  ;  5  X  G  =  30  feet. 

The  spcciiication  is  made  out  according  to  the  last  solution. 

Specif  nation  of  Batten  Shingle,  No.  12. 


n 


o 

t^ 

00 

C5 

o 

1^ 

a  m 
Lengths,  g  § 

5" 

X 

X 

X 

X 

X 

—I'M 

2 

X 

X 

IN 

4 

Contents. 

20 

45 

15 

8 

4 

12 

2,812 

21 

8 

3 

4 

4 

3 

4 

3 

1.080 

22 

9 

4 

3 

3 

2 

1 

3 

917 

23 

4 

3 

4 

4 

3 

3 

4 

1.073 

24 

12 

8 

4 

2 

1 

4 

12 

1.880 

25 

6 

9 

4 

3 

3 

3 

2 

1,276 

26 

24 

8 

4 

3 

2 

1 

3 

1,7G5 

Total,  10,803 

Random  Shingle  No.  13,  for  any  Dimension. 
{Contents  given  in  the  Columns.) 


a  £ 

100 
200 
2o() 
300 
275 
8G8 
210 
450 
150 

CO 

X 

700 
120 
160 
240 
100 
120 
200 
250 

X 

250 
210 
640 
120 
240 
I.tO 
180 
200 
160 

CO 

X 

CI 

150 

250 
300 
350 
4<»0 
500 
120 
240 
60 

CI 

r— t 

X 

120 
210 
150 
320 
150 
210 
641 
120 

X 

CO 

120 
100 
100 
210 
120 
250 
100 
100 

X 

CO 

150 
120 
(iO 
20 
40 
36 
12 

1— c 

X 

250 
160 
500 
250 

00 

X 

210 

420 
150 
600 
500 
120 

? 

210 
410 
210 
312 
200 
100 
100 

1-1 

X 

O 
rH 

120 
250 
120 
120 
200 
120 
100 

2 
X 

ox 

I-l 

120 
120 
600 
150 
120 
210 
100 

CO 

X 

160 
120 
150 
100 
200 
150 
200 
150 

C5 

X 

200 
200 

2r,() 

100 
100 
100 
100 
200 

rH 

X 

120 
210 
l.'iO 
120 
150 
100 
200 
120 

46 


SKLF-INSTRUCTOR 


I  IMS! 


iiilii 


Method  of  keeping  Shingle  No.  13. 

The  contents  are  found  by  the  Board  Rule  and  marked 
on  each  piece,  and  afterwards  placed  in  the  proper  column  in 
the  shingle. 

What  is  the  total  number  of  feet  of  merchantable  spruce 
lumber  in  Random  Shingle,  No.  13.  Ans.  23,464  feet. 

Random  Shingle,  No.  14. 
[Running  Lengtlis  given  in  the  Columns.) 


s 

to 

00 

<s> 

la 

« 

tt> 

t- 

O) 

s 

Contents  of  the 

X 

X 

X 

X 

X 

X 

X 

X 

X 

X 

X 

wliole. 

100 

CO 

80 

100 

200 

72 

120 

to 
100 

l- 

120 

I- 

20 

00 
120 

O 

60 

3X   6=  1,045 

100 

6'l 

150 

100 

72 

lOi) 

10() 

100 

18 

150 

20 

2X10=  2,833 

25 

40 

210 

75 

60 

150 

120 

100 

16 

120 

40 

4X  3=   4,186 

125 

20 

110 

60 

40 

100 

2ii0 

40 

24 

100 

20' 

4X  'J=  2,970 

IW 

15 

200 

40 

18 

110 

110 

20 

20 

100 

100; 

5X  6=    1,096 

2(10 

- 

_ 

20 

19 

70 

120 

loo 

18 

200 

lOOi 

6X  «=  2,137 

KK) 

28 

150 

10 

20 

60 

150 

60 

10 

100 

100 

6X  6=  3-525 

100 

30 

120 

75 

70 

40 

loo 

60 

19 

150 

_ 

7X  7=  2,797 

100 

72 

100 

100 

60 

20 

60 

40 

24 

250 

150 

7X  9=   1,118 

200 

150 

12' 

150 

40 

30 

40 

20 

20 

loo 

200 

8X10=10,466 

150 

9f5 

150 

liO 

20 

20 

20 

15 

- 

120 

i2o: 

10X12=10,600 

400 

100 

160 

100 

15 

20 

25 

20 

18 

_ 

_  1 

1700 

697 

1570 

990 

20 
526 

16 

855 

3a 

1175 

685 

213 

60 

150 
1050 

1570 

n 

n 

•if, 

3 

2] 

2,^, 

3 

4A 

5i 

61 

10 

Total   .  42,673  ft. 

1700 

(597 

3140 

2970 

1052 

1710 

3525 

2740 

106.5 

9420 

10500 

1133 

318 

1046 

44 

427 

57 

53 

1046 

2833 

1U45 

4186 

1096 

2137 

2797 

1118 

10463 

ON  LUMBER   SURVEYING. 


47 


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48 


SELF-INSTRUCTOR 


Jinle  shotviiig  how  Table  B  is  calculated. 
Divide  the  area  or  contents  of  the  end  into  the  given 
number  of  feet  of  contents,  and  the  quotient  will  be  the 
number  of  feet  of  running  lengths,  equivalent  to  the  given 
number  of  feet  of  contents. 

1.  What  number  of  feet  in  length  of  10  inches  by  12 
inches  will  be  equal  to  1,000  feet  contents. 

By  the  table  10  inches  X  12  inches  is  10  times  the  length, 
for  the  contents ;  therefore,  l,000-^10=100  feet  in  length. 

2.  How  many  feet  of  2  X  3  are  equal  to  1,000  feet  of 
contents  ? 

2  X  3  =  i  the  length ;    therefore,    1,000  X  2  =  2,000 
feet  =  length  required. 

Table  C. — Number  of  Feet  of  the  folUncing  Dimensions  of  Timber 
that  will  make  1,000  Feet,  Cubic  or  Solid  Measurement. 


DimensiuDS. 

No.  of  Feet 
iu  Lengtli. 

Cubic 
Feet. 

Dimensions. 

No.  of  Feet 
iu  Length. 

No.  of  ft. 
of  Cubic 
Measure. 

5   X 

5 

5,760 

1,000 

7  X  12 

l,714f 

1,000 

5  X 

6 

4,800 

1,000 

8X8 

2,250 

1,000 

5  X 

7 

4,114^ 

1,000 

8X9 

2,000 

1,000 

5  X 

8 

3,600 

1,000 

8  X  10 

1,800 

1,000 

5  X 

9 

3,200 

1,000 

8  X  11 

1,636t4j- 

1,000 

5  X 

10 

2  880 

1,000 

8  X  12 

1.500 

1,000 

5  X 

11 

2,618T2r 

1,000 

9X9 

1,7775 

1,000 

5  X 

12 

2,400 

1,000 

9  X  10 

1,600 

1,000 

6  X 

6 

4,000 

1,000 

9  X  11 

1,455^^ 

1,000 

6  X 

7 

3,428f 

1,000 

9  X  12 

1,333^ 

1,000 

6  X 

8 

3,000 

1,000 

10  X  10 

1,440 

1,000 

6   X 

9 

2,666f 

1,000 

10  X  11 

l,309yV 

1,000 

6  X 

10 

2,400 

1,000 

10  X  12 

1,200 

1.000 

6  X 

11 

2,181^^ 

1,000 

11  X  11 

l,190i\\ 

1,000 

6  X 

12 

2,000 

1,000 

11  X  12 

1,090}^ 

1,000 

7  X 

7 

2,9383| 

1,000 

12  X  12 

1,000 

1,000 

7  X 

8 

2,571^ 

1,000 

14  X  16 

642^ 

1,000 

7  X 

9 

2.285f 

1,000 

16  X  18 

500 

1,000 

7  X 

10 

2,057| 

1,000 

18  X  20 

400 

1,000 

7X 

11 

l,870|f 

1,000 

20  X  22 

327t\ 

1,000 

" 

^ 

1,000 

22  X  24 

272i\ 

1,000 

ON   LUMBER   SURVEYING. 


49 


Hule  showing  how  Table  C  is  computed. 

Multiply  the  breadth  and  width  in  inches  together,  and 
divide  the  product  by  144,  the  number  of  inches  in  a  square 
foot,  and  the  quotient  divided  into  the  given  number  of  cubic 
feet  will  give  the  number  of  feet  in  length,  ec^ual  to  said 
number  of  feet. 

How  many  feet  running  length  of  6  inches  X  G  inches  are 
equal  to  1,000  cubic  feet?  J71S.  4,000  feet. 

6  X  6  =  36  ;  36  -4-  144  =  ^Y*  =  i  ;  i  inverted  =  to  t 
X  U)^"-«  =  -tOjai)  =  4,000  feet  of  running  lengths  =  1 ,000 
cubic  feet. 


1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1.000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 

1,000 


Table  showing  the  Numbers  to  multiply  the  Lengths  of  the 
following  Dimensions   by  in   order  to  find  the   Contents 
in  Cubic  Feet. 


Dimension. 

No. 

Dimension. 

No. 

Dimension. 

N'o. 

5  X     5 

-* 

7  X 

11  — 

/?V 

12  X 

16  -r: 

H 

5  X     6 

-A 

7  X 

12  — 

^h 

13  X 

14  ^- 

in 

5X7 

m 

8  X 

8  — 

^ 

14  X 

16  ^-. 

1^ 

5X8 

5 
15 

8  X 

9  — 

1 
2 

16  X 

18  — 

2 

6X9 

-tv 

8  X 

10 

tl 

16  X 

20  ^- 

2f 

5  X  10 

n 

8  X 

11  — 

\\ 

18  X 

20  rrr 

H 

5  X  11 

-  tV* 

8  X 

12  — 

2 
3 

20  X 

22  — 

3t3 

5  X  12 

-tV 

9  X 

9  — 

Iff 

22  X 

2t  rr-. 

H 

6  X     6 

—  JL 
4 

9  X 

10  — 

5 

8 

24  X 

26  = 

4^ 

6  X     7 

-/t 

9  X 

11  — 

H 

26  X 

28  — 

5tV 

6X8 

—  A 

.....  ^ 

9  X 

12 

4 

28  X 

30  —. 

H 

6  X     9 

-f 

10  X 

10  — 

u 

30  X 

32  — 

6f 

6  X  10 

-tV 

10  X 

11 

H 

32  X 

34  — 

7^ 

6  X  11 

-  h\ 

10  X 

12  — 

34  X 

36  — 

8i 

6  X  12 

i 

11  X 

11 

m 

36  X 

38  r^ 

9^ 

7X     7 

=  tV? 

11  X 

12  = 

H 

38  X 

40  —. 

10i{ 

7  X     8 

-■/tt 

12  X 

12  — 

1 

40  X 

42  — 

111 

7X9 

=  A 

12  X 

14  — 

H 

42  X 

44  = 

12^ 

7X  10 

-n 

4 


'I     < 


SKLF-INSTRUCTOR 


QUKSTI0N3    FOK    KXKUCISK. 

1.  Required  the  nutnl)er  of  solid  feet  in  a  timber  G  inches 
X  ^'  inches  and  40  feet  long?  A)is.  10  feet. 

Solution.  —  G  X  C  =  ]  of  length,  therefore  \  of  40  = 
10  feet. 

2.  AVhat  is  the  solidity  of  a  j>iece  of  6-inch  X  12-iii('h 
timber  72  feet  long  ?  Ans.  3G  feet. 

15y  the  table  G  X  12=  2  ^^^  length  ,  the  contents; 
therefore  J-  X  72  =  M)  feet. 

3.  What  number  of  cubic  feet  are  there  in  a  piece  of 
timber  40  I'eet  long,  22  inches  X  ^'1  inches  ? 

Ans.  146§  feet. 

4.  Reqiiirei?  the  number  of  feet  in  a  piece  of  timber  32 
feet  long,  5  incl'es  X  1-  inches?  Ans.  13^  feet. 

Solution.  —  ^2  feet  X  tV  =13^  feet  =  contents. 

5.  What  number  of  cubic  feet  in  the  following  pieces, 
namely,  6  pieces  GO  feet  long  12  inches  X  IC  inches,  and  12 
pieces  35  feet  long  and  IG  inches  X  1^  inches? 

\ns.  15,840  feet. 

6.  What  are  the  contents  in  cubic  f  "  G  pieces  of  20 
inches  X  '^4  inches  and  35  feet  long  ? 

Ans.  1 1 1  §  cubic  feet. 

7.  What  number  of  cubic  feet  in  a  piece  of  timber  28 
inches  X  30  inches  and  GO  feet  long?     Ans.  350  cubic  feet. 

Solution.  —  GO  X  ^ij  =;  350  feet  of  cubic  measm-e. 

8.  Required  the  contents  in  cubic  feet  of  a  piece  of  pine 
timber  30  inches  X  32  inches  and  30  feet  in  length  ? 

Ans.  200  feet. 

9.  ITow  many  tons  of  timber  (allowing  42  cubic  feet  to 
the  ton)  in  a  piece  of  timber  38  inches  X  40  inches  and  45 
feet  long?  Ans.  11|^  tons. 

10.  What  will  be  the  cost  of  a  piece  of  pine  timber  18 
inches  X  20  inches  and  30  feet  in  length  @  30  cents  per 
cubic  foot  ?  Ans.  $22.50. 


ON  LUMBER   SUHVEYING. 


51 


nchcs 
)  feet. 
40  = 

2-iiK',li 
G  i'eet. 
ileiitH ; 


ece 


of 


6§  feet, 
fiber  32 
3^  feet. 


pieces, 
,  and  12 

B40  feet, 
es  of  20 

il)ic  feet. 
Imber  28 
iil)ic  feet. 
|e. 
of  pine 

>00  feet. 

feet  to 

and  45 

14  tons. 

imber  18 

tents  per 

$22.50. 


Rule  to  reduce  Feet  of  Board  Measure  to   Cubic  Feet. 

Divide  the  contents  in  snperficiul  feet  by  12,  and  it  will 
give  the  number  of  cubic  feet;  or  multiply  the  number  of 
cubic  I'eet  by  12  and  tlie  product  will  be  feet  of  board 
measure. 

In  1,200  feet  of  board  measure  how  many  cubic  feet  are 
there  ?  Ans.  100  cubic  feet. 

Solution.  —  1,200  -^  12  =  100  cubic  feet. 

llequired  the  number  of  feet  of  board  measure  in  100 
feet  of  cubic  measure  ?  Ans.  1,200  feet. 

100  X  12  =  1,200  feet  of  board  measure. 

Second  Method  of  malcing  out  a  Specification. 

3-INCII    SriiClFICATION    BY    THE    SECOND    METHOD. 


1 

2  "■ 

» 

t- 

QO 

O) 

o" 

Lengths,    g  § 

X 

X 

X 

X 

05 

X 

X 

X 

Contents. 

14 

2 

3 

4 

6 

8 

4 

6 

15 

4 

2 

1 

4 

2 

8 

4 

16 

2 

4 

2 

1 

3 

2 

4 

17 

« 

1 

1 

3 

2 

18 

8 

4 

6 

1 

3 

2 

4 

19 

2 

1 

2 

3 

2 

4 

6 

20 

3 

2 

1 

4 

2 

1 

3 

21 

6 

4 

8 

2 

1 

3 

2 

22 

1 

5 

4 

3 

2 

1 

1 

23 

2 

1 

10 

4 

1 

2 

1 

24 

6 

4 

3 

2 

25 

4 

2 

8 

6 

4 

26 

3 

2 

1 

2 

8 

27 

6 

5 

1 

3 

2 

28 

8 

2 

4 

6 

29 

3 

5 

2 

1 

6 

4 

1510 

1864 

2092 

2140 

1717 

2563 

3807 

15,693  feet. 

ffir 


.     1 


'W 


I      i 


52 


SELF-INSTRUCTOR 


Second  Hulefor  Specifications.    . 

Multiply  the  number  of  pieces  or  dots  in  each  square  of 
the  specification  by  the  length  of  one  of  the  pieces ;  and 
multiply  the  product  thus  found  by  ^  of  the  breadth  of  said 
pieces  for  the  contents  in  board  measure  of  3-inch  deals;  by 
^  of  the  breadth  lor  4-inch  ;  by  g^  of  it  for  plank,  etc. 

Example  showing  how  to  make  out  the  Three-inch  Speciji- 
cation  hy  Second  Method. 


First  Column  6  inches  wide. 

Second  Column  7  inches  wide. 

14 

X  2  —    23 

14  X 

3  —     42 

15 

X  4           60 

15  X 

?  —     30 

16 

X  2  —    32 

16  X 

4  —    64 

17 

X  6  —  102 

18  X 

4  —     72 

18 

X  8         144 

19  X 

1  —     19 

19 

X  2  —    38 

20  X 

2  —    40 

20 

X  3  —    60 

21  X 

4           84 

21 

X  6  =  126 

22  X 

5  —  110 

22 

X   1           22 

23  X 

1  =    23 

23 

X  2—    46 

24  X 

6         144 

25 

X  4  —  100 

26  X 

3  —    78 

27 

X  6         162 

27  X 

5  —  135 

29 

X  3           87 

28  X 

8  —  224 

1,007 

1,065 

4 

If 

1,007 

1,065 

503 

Conte 

799 

Contents,  1,510  feet. 

nts,  1,864  feet. 

6  inches 

,  the  breadth,  di- 

7  inches, 

the  breadth,  di- 

vided  by  • 

±  is       to  li,  and 

vided  by  4 

is       to  1|,  and 

U  X   1,007  =  1,510,  the 

If  X  1,065: 

=  1,864  feet  = 

contents. 

contents. 

English  deal  specifications  are  generally  made  out  by  the 
second  method.     Both  rules  will  give  the  same  results. 


ON   LUMBER   SURVEYING.  53 

Specification  of  Philadelphia  Deal  Shingle. 


C   00               r-l 

.                  (N 

Lengths.    1  § 

5  * 

X 

Contents. 

Iiengths.   g  §        y 

Contents. 

eo 

5  " 

M 

14 

40 

1,680 

28 

14 

1.176 

16                 35 

1,680 

30 

8              720 

18                 30 

1,620 

32 

4               384 

20                 11 

660 

34 

4              408 

22 

9 

594 

36 

7              756 

24 

21 

1,512 

38 

14            1,596 

26 

6 

468 

40                14            1,680 

Contents,  14.934  feet. 

Philadelphia 

Deal  Shingle. 

a  00 
Lengths,    g  g 

5"^ 

l-H 

X 

1      , 

Lengths.    |§ 

2" 

X 

eo 

H 

M 

14 

40 

28                  

U 

16 

30 

36 

8 

18 

32 

•  ••• 

30 

4 

20 

34 

•  «•• 

11 

4 

22 

9 

36 

7 

24 

21 

38 

14 

40 

26 

6 

14 

-  _j  irmjdMgMai^M^B 


64 


SELF-INSTRUCTOR 


IM 


I! 


The  specification  of  Philadelphia  deals  is  done  the  sanic 
as  the  3-incli  specification ;  or  multiply  the  running  lengths 
by  3  for  the  contents  in  fi^et  of  board  measure.  Philadel- 
phia deal  is  generally  12  inches  wide  and  even  lengths,  from 
1 4  feet  up,  and  tlie  best  quality  of  spruce  lumber.  English 
deals  generally  comprise  all  deals  too  short,  or  not  good 
enough  for  Philadelphia  or  New  York  deals.  Also  short 
timber,  battens,  and  plank,  not  suitable  for  other  markets,  go 
into  the  English  deal  pile.  Deals  that  are  knotty,  cracked 
by  the  sun,  or  stained,  or  having  wanes  on  them,  and  not 
poor  enough  for  refuse,  go  to  the  English  deal  pile.  New 
York  deal  must  be  the  best  quality  of  spruce,  from  14  feet 
long  up. 

Directions  showing  how  to  measure  all  hinds  of  Lumber 

by  the  Board  Rule. 

Lay  your  rule  across  the  board  to  be  measured,  at  right 
angles  to  the  further  edge  of  the  board,  and  let  the  outside 
edge  of  the  board  and  further  end  of  the  rule  be  both  even 
on  that  side,  then  observe  the  length  of  your  board  and  turn 
your  rule  to  the  same  length,  then  look  on  the  line  or  column 
of  that  length,  and  you  will  find  the  contents  marked  on  the 
rule  just  over  the  inside  edge  of  the  board. 

EXAMPLES    FOR    PRACTICE. 

1.  What  are  the  contents  of  a  1^-inch  board  IG  feet  long 
and  12  inches  wide?  Ans.  20  feet. 

By  the  rule  the  contents  given  for  1-inch  board  is  IG  feet 
contents,  to  which  add  ^  of  the  contents,  which  will  give 
the  contents  for  1^ -inch  boards.  16-4-4=4;  16-|-4  = 
20  feet  contents. 

2.  What  are  the  contents  of  a  board  32  feet  long  and  12 
mches  wide  ?  Ans.  32  feet. 

As  there  is  no  32  on  my  rule,  I  find  the  contents  by  the 
rule  of  a  board,  half  the  length  to  be  IG  feet;  which  being 
doubled,  gives  the  contents  required  =  32  feet. 

3.  What  are  the  contents  of  a  1^-itich  board  20  feet  long 


and  12  inches  wide? 


Ans.  30  feet. 


e  saniG 
engths 
liladel- 
s,  from 
^^nglish 
t  good 
3  short 
kets,  go 
snicked 
md  not 
,     New 
14  feet 

Lumber 

at  right 
outside 

)th  even 

md  turn 
column 

1  on  the 


eet  long 
20  feet. 
16  feet 

l^ill  give 

+  4  = 

and  12 
32  feet, 
by  the 
;h  being 

Tfeet  long 
30  feet. 


ON  LUMBER   SURVEYING. 


55 


By  the  rule  an  inch  board  20  feet  long  and  12  inches 
wide  will  contain  20  feet,  to  which  add  half  of  20  for  the 
contents  of  a  l^-inch  board.  20  -f-  2  =  10 ;  20  +  10  =  30 
feet. 

4.  Required  the  contents  of  a  plank  24  feet  long  2  inches 
X  12  inches?  Ans.  48  feet. 

By  the  board  rule,  in  a  board  24  feet  long  12  inches  wide 
and  1  inch  thick  there  are  24  feet,  and  as  plank  is  2  inches 
thick,  therefore  twice  the  contents  of  the  face  of  it  will  be 
equal  to  the  true  contents,  24  X  2  =  48  feet. 

Rule  for  any  Dimension. 

Multiply  the  number  of  feet  in  the  face  of  the  piece  to 
be  measured,  by  the  thickness  in  inches,  and  it  will  give  the 
contents  in  feet  of  board  measure. 

Rule  for  measuring  Logs  or  Round  Timber. 

Multiply  the  length,  taken  in  feet,  by  the  square  of  one 
fourth  of  the  mean  girth,  taken  in  inches,  and  this  product 
divided  by  1 44  will  give  the  contents  in  cubic  feet. 

Note.  —  The  girth  of  tapering  timber  is  usually  taken 
about  one  third  the  distance  from  the  larger  to  the  smaller 
end.  The  rule  is  that  in  common  use,  though  very  far  from 
giving  the  actual  number  of  cubic  feet ;  40  cubic  feet  as 
given  by  the  rule  are  in  fact  r:=  oO^^q  true  cubic  feet. 

EXAMPLE. 

1.  How  many  cubic  feet  in  a  stick  of  timber  which  is  40 
feet  long,  and  whose  girth  is  GO  inches  ?  Ans.  62^  feet. 

GO  -f-  4  =  15  inches  rr=  l  of  girth  ;  15  X  !•'>  =  225  ~ 
square  of  quarter  of  the  girth ;  225  X  'iO  feet  =  'J,000 ; 
9,000  -f-  144=  G2^  cubic  feet. 

2.  How  many  cubic  feet  in  a  piece  of  timber  21  feet  long, 
and  whose  girth  is  36  inches  ? 

3.  What  are  the  contents  of  a  log  100  feet  long,  and 
whose  tiirth  is  150  inches? 


''!:f=« 


I    -l 


56 


SELF-INSTRUCTOR 


To  jind  the  largest  Square  Piece  of   Timber  that  may  he 
sawed  from  a  Round  Stick  of  Timber,  having  the  Diam- 
eter or   Circumference  of  the  Small  End  given. 
Ride  1.  —  Multiply  the  given  diameter  by  .707106,  or, 

multiply  the  given  circumference  by  .225079.     Or,  as  the 

diameter  of  a  circle  is  equal  to  the  diagonal  of  the  inscribed 

square  — 

Rule  2.  —  Square   the   diameter 

and  take  half  the  sum  of  the  i^quare,  I 

and  extract  the  square  root  of  it, 

ui.d  the  root  thus  found  will  be  the  side  of  the  inscribed 

square. 

EXAMPLE. 

1.  I  have  a  piece  of  timber  30  inches  in  diameter ;  how 
large  a  square  stick  can  be  hewn  from  it. 

By  the  last  rule  30  squared  =  30  X  30  =  900 ;  900-7-2 
=  450;  ^5!^  ==  21.21 -{-inches  square. 

2.  How  large  a  square  stick  may  be  hewn  from  a  piece 
of  round  timber  120  inches  in  circumference? 

3.  How  large  a  square  stick  may  be  sawn  from  a  piece 
of  round  timber  60  inches  in  diameter  ? 

Having  the  Side  of  a  Square  Stick  given^to  find  the  Diam- 
eter of  the   Tree  from  which  it  was  sawn. 

Rule.  —  Square  the  side  and  double  it,  and  out  of  the 
product  extract  the  square  root. 

What  must  be  the  diameter  of  a  tree  that  when  hewn 
shall  be  18  inches  square?  Ans.  25.44 inches. 

Table. 
12  lines  =  1  inch. 
12  inches  =  1  foot. 
3  feet  =  1  yard. 

Inches  multiplied  by  inches  produce 
Parts  marked  thus  '. 
Parts  by  parts  give  fourths,  marked  thus  ''". 


ON  LUMBER   SURVEYING. 


57 


Inches  are  marked  '. 
144  square  inches  make  1  square  foot. 
9  square  feet  :^  1  square  yard. 
1,728  cubic  inches  =  1  cubic  foot. 
50  cubic  feet  =  1  load. 
40  cubic  feet=  1  ton  of  timber. 
1 6  cubic  feet  =  1  cord  foot. 
8  cord  feet,  or  1 28  cubic  feet  =  1  cord  of  wood. 
1,980  feet  superficial  =  1   St.   Petersburg   standard    of 
deals. 

Form  of  a  Bill  of  Lading  of  Timber,  Shingle  No.  8,  etc.,  etc. 

Shipped,  in  good  order  and  condition,  by  Edmond  B. 
Sanderson  &  Co.,  on  board  the  good  ship  "  Southern," 
whereof  James  Brown  is  master  for  this  present  voyage, 
now  lying  in  the  port  of  New  York,  U.  S.,  and  bound  lor 
Liverpool,  Englp'id.     To  say:  — 

47,928  ft.  INIer.  spi-uce,  all  under  deck, 
100  M  spruce  laths,  all  mider  deck, 
80  M  ft.  Mer.  pine,  all  on  deck, 

being  marked  and  numbered  as  in  the  margin ;  and  are  to 
be  delivered,  in  like  good  order  and  condition,  at  the  afore- 
said port  of  Liverpool  (the  danger  of  the  seas  and  fire 
always  excepted),  unto  David  Belt  &,  Sons,  or  to  assigns, 
he  or  they  paying  freight  for  the  said  timber  at  the  rate  of 
ten  dollars  per  M  feet,  and  one  dollar  per  M  for  laths,  with- 
out primage  and  average  accustomed. 

In  witness  whereof  the  master  of  the  said  vessel  hath 
affirmed  to  three  bills  of  lading,  all  of  this  tenor  and  date  ; 
one  of  which  being  accomplished,  the  others  to  stand  void. 

James  Brown. 

Dated  at  New  YonK,  U.  S., 
May  the  3d,  A,  D.  1870. 


inn 


58 


SELF-INSTRUCTOR 


Bill  of  Lading. 


il        ! 


SiiiprKD,  in  good  order  and  condition,  by  T. 
Pandul  &  Co.,  on  board  the  good  schooner  culled 
the  "  Northern  Dawn,"  whereof  Daniel  K.  Bloomer 
is  master  for  this  present  voyage,  now  lying  in  the 
port  of  Bangor,  Me.,  and  bound  for  New  York. 
To  say :  — 


S2 

'o 

V 

u 

c 
s 

<3i 

G 
O 

6 
'H 

s 


> 
o 
x> 
cS 

a> 

e 
o 

12 

"5 

Ph 


1 1 0  M  feet  hemlock  lumber,  all  vnider  deck, 
75  M  feet  spruce  lumber,  all  on  deck, 
120  M  laths,  all  on  deck, 

being  marked  and  numbered  as  in  the  margin ;  and 
are  to  be  delivered,  in  like  good  order  and  condition, 
at  the  aforesaid  port  of  New  York  (the  danger  of 
the  seas  and  fire  only  excepted),  unto  Messrs.  Den- 
ton and  Beeters,  or  to  assigns,  he  or  they  paying 
freight  for  the  said  lumber  at  the  rate  of  four  dol- 
lars per  M  feet,  and  sixty  cents  per  M  for  laths, 
without  primage  and  average  accustomed. 

In  ivitness  whereof,  the  master  of  the  said  vessel 
hath  affirmed  to  three  bills  of  lading,  all  of  this 
tenor  and  date  ;  one  of  which  being  accomplished, 
the  others  to  stand  void. 

Daniel  E.  Bloomer. 

Dated  at  Bangor,  Me., 

June,  the  3d,  18G9. 


Surveyor's  Bill  for  Services  rendered. 

Bangor,  Me.,  June  the  2d,  1809. 

Messrs.  DuNTON  &  Boomer, 

To  Daniel  E.  Siiaw,  surveyor.  Dr. 
For   surveying   250    M   ft.  of  spruce   lumber  to 

schooner  "  Juno,"  @  25c.  per  M $62.50 


ON  LUMBER   SURVEYING. 


59 


Survey  Bill  of  Lumber,  etc. 

Surveyed  from  James  E.  D<alo  &  Sons,  of  Clinton,  Iowa,  to 
schooner  "  Pallas,"  Captain  Dunn.      To  say  :  — 

3G,o00  ft.  2  X  6,  from  12  ft.  long  up  (mch.),  spruce. 
35,G00  «  No.  1  pine  boards. 
22,400  "  hemlock  Ijoards  (mch.). 
15,000  "  8  X  10  Mer.  pine  timber. 
250  M  No.  1  pine  shingles. 

Thomas  B.  Proudfoot, 

Surveyor. 
Clinton,  Iowa,  ^ 

June  the  12th,  Anuo  Domini  1869. 

Surveyor's  Receipt. 

^"^•^^*  Bangor,  Me.,  June  the  4th,  A.D.  1869. 

Received  from  Messrs.  Dunton  &  Boomkr  sixty-two 
dollars  and  fifty  cents,  which  pays  for  surveying  250  M  feet 
of  spruce  lumber  to  schooner  "  Juno,"  @  25c.  per  M. 

Daniel  E.  Shaw,  Surveyor. 


NOVEL   RULES 

For  finding  the  Contents  of  Planh,  Deal,  Battens,  Joist,  and 
Timber,  by  multiplying  a  Fractional  Part  of  the  Length 
by  the  Breadth. 

2-inch  is  ^  of  the  length  multiplied  by  the  breadth,  for  the 
contents. 

3-inch  is  I  of  the  length  multiplied  by  the  breadth,  for  the 
contents. 

4-inch  is  ^  of  the  length  multiplied  by  the  breadth,  for  the 
contents. 

5-inch  is  the  length  divided  by  2f ,  and  the  quotient  multi- 
plied by  the  breadth. 


BBnn 


fi 


60 


SELF-INSTRUCTOR 


ii 


6-inch  is  ^  of  the  length  multiplied  by  the  breadth,  for  the 
contents. 

7-inch  is  the  length  divided  by  1^,  and  the  quotient  multi- 
plied by  the  breadth. 

8-inch  is  the  length  divided  by  1^,  and  the  quotient  multi- 
plied by  the  breadth. 

9-inch  is  the  length  divided  by  IJ,  and  the  quotient  multi- 
plied by  the  breadth. 
10-inch  is  the  length  divided  by  1^,  and  the  quotient  multi- 
plied by  the  breadth. 
11-inch  is  the  length  divided  by  1^^,  and  the  quotient  multi- 
plied by  the  breadth. 
12-inch,  multiply  the  length  by  the  width,  for  the  contents. 

2^-inch,  or  battens,  is  the  lengtii  divided  by  4f,  and  the 
quotient  multiplied  by  the  breadth. 

P.  S.  —  The  above  rules  give  the  contents  in  feet  of  board  measure. 
EXAMPLES    FOR   PRACTICE. 

1.  Required  the  contents  in  superficial  feet  of  a  piece  of 
timber  10  inches  X  12  inches  and  40  feet  long. 

Jns.  400  feet. 
Solution.  —  By  the  table,  10  inches  is  11  of  the  length  mul- 
tiplied by  the  breadth.     Therefore  40  feet  -4-  1^  =  Y-  X 
^  =  200  =  33^  ;  33^  X  12  =  400  feet. 

2.  What  are  the  contents  of  a  piece  of  timber  12  inches 
X  20  inches,  and  40  feet  long  ?  Ans.  800  feet. 

Solution.  —  40  X  20  =  800  feet. 

3.  What  are  the  contents  of  a  plank  2  inches  X  1 1  inches 
and  36  feet  long?  Ans.  G6  feet. 

Solution.  —  2   inches   is   J   of  the    length.      Therefore 
36-f-6=6;  6X11  =  66  feet. 

4.  What  are  the  contents  of  a  piece  of  timber  8  inches  X 
11  inches  and  40  feet  in  length?  Ans.  293^  feet. 

Solution.  —  40  -^  1^  =  -Y- ;  *r-  X  §  =  V'  =  26§  ;  26§ 
X  ll  =  293ifeet. 


ON  LUMBER   SURVEYING. 


Gl 


Given  the  Breadth  of  a  Rectancfular  Plank  in  Inches,  to 
find  how  much  in  Length  will  make  a  Foot^  or  any  other 
required  Quantity. 

Ride.  —  Divide  1 44,  or  the  area  to  be  cut  off,  by  the 
breadth  in  inches,  and  the  quotient  will  be  the  length  hi 
inclies. 

1.  If  a  boaid  be  6  inches  broad,  what  length  of  it  will 
make  a  square  foot  ?  Ans.  2  feet. 

Solution. —  144  inches  -i-  6  inches  =  24  inches;  24 
inches  -f-  12  inches  =  2  feet. 

2.  If  a  plank  be  2  inches  X  8  inches  in  size,  what  length 
of  it  will  make  4  square  feet  ?  Ans.  3  feet. 

Solution.  —  2  X  8  =:  1 0,  area  of  the  end  ;  1 44  -^  1 G  =  9 
inches  for  1  foot,  which,  being  multiplied  by  4  =  4  X  9  = 
36  inches  =  3  feet. 


To  find  the  Solid  Contents  of  a  Piece  of  Timber  tapering 

regularly. 

Ride.  —  Multiply  the  sum  of  the  breadths  of  the  two  ends 
by  the  sum  of  the  de[)ths,  to  which  add  the  product  of  the 
breadth  and  depth  of  each  end ;  ^  of  this  sum,  multiplied  by 
the  length,  will  give  the  exact  solidity  of  any  piece  of 
squared  timber  tapering  regularly. 

1.  How  many  feet  in  a  piece  of  mahogany  whose  ends 
are  rectangles,  the  length  and  breadth  of  one  being  14  and 
12  inches,  and  the  corresponding  dimensions  of  the  other 
end  6  and  4  inches ;  also  the  length  30^  feet  ? 

Ans.  18^  cubic  feet. 
Solution.  — 

14_|_G  =  20  12X14  =  168 

12-1-4=16  6  X    4=    24 

20  X  16  =  320 

512  sq.  in.  =  A^  sq.  ft. 
Then  |  X  ¥  X  30^  =  18^  cubic  feet. 


m 


62 


SKLF-INSTRUCTOR 


When  a  Board  or  Plank  is  broader  at  one  End  than  the 
other,  to  find  what  Length  of  it  will  make  a  Foot,  or 
any  other  rcqtnred  Quantity. 

Rule.  —  To  the  square  of  the  product  of  the  length  and 
narrow  end  add  twice  the  continual  product  of  tiicse  (juan- 
tities  ;  namely,  the  length,  the  diiference  between  the  breadths 
of  the  ends,  and  the  area  of  the  part  re(|uired  to  be  cut  off. 
Extract  the  square  root  of  the  sum  ;  from  the  result  deduct 
the  product  of  the  length  and  narrow  (.nd,  and  divide  the 
remainder  by  the  difference  between  the  breadths  of  the 
ends. 


EXAMPLE.  j(^ 

It  is  required  to  cut  off  GO  inches 
from   the   smaller   end    of  a    board ;  8 
A  D  being  3  inches,  C  E  6  inches, 
and  A  B  20  inches. 


Here  Aa:=Jyn3(v{(^^XADV+4BCx 
A  B  X  60|  — A  B  X  A  D  =^^^1^20  X  3\'  + 
6  X  20  X  60  I  —  20  X  3  =  14.64,  the  lengtli  required. 

To  find  hoio  much  in  Length  will  make  a  Solid  Foot,  or  any 
other  required  Quanfify,  of  Squared  Timber,  of  equal 
Dimensions  from  End  to  End. 

Mule.  —  Divide  1,728  —  the  solid  inches  in  a  foot,  or  the 
solidity  to  be  cut  off — by  the  area  of  the  end  in  inches. 

1.  If  a  piece  of  timber  be  14  inches  broad  and  10  inches 
deep,  how  much  of  it  will  make  a  solid  foot  ? 

Ans.  12^1  inches,  the  length  required. 

10  X  14  =  140  ;  1,728  ~  140  =  12if  mches. 


ON   LUMBER   SURVEYING. 


68 


Bide.  —  Multiply  tlio  area  corresponding  to  the  quarter 
girt  in  inches,  hy  the  length  of  the  piece  in  feet,  and  the 
product  will  he  the  j?olidity.  If  the  (piarter  girt  exceeds  the 
Hunts  of  the  table,  take  ^  of  it,  and  4  times  the  contents  thus 
found  will  give  the  re([uired  contents. 

A    Table  for  Measuring   Timber. 


Quarter  Girt. 

Area. 

Quarter  Girt. 

Area. 

Quarter  Girt. 

Area. 

Inches. 

Feet. 

Inrht'B. 

Feet. 

1 

Inches. 

Fret. 

6 

.250 

12 

1.000 

18 

2.250 

1 

.272 

12i 

1.042 

18i 

2.376 

294 

121 

1.085 

19 

2.506 

^i 

.317 

12} 

1.129 

19^ 

2.640 

7 

.340 

13 

1.174 

i         20 

2.777 

7i 

.364 

13: 

1.219 

1         20i 

2.917 

7^ 

.390 

1.3;  r 

1.265 

21 

3.062 

7^ 

.417 

isl 

1.313 

211 

3.209 

8 

.444 

14 

1.361 

22 

3.362 

8i 

.472 

14i 

1.410 

22^ 

3.516 

8^ 

.501 

l4 

1.460 

i        23" 

3.673 

8j 

.531 

14^ 

1.511 

231 

3.835 

9 

.562 

15 

1.562 

24 

4.000 

9i 

.594 

15i 

1.615 

1    241 

4.168 

9i 

.626 

15i 

1.668 

25 

4.340 

9f 

.6.59 

15^ 

1.722 

:    25i 

4.516 

10 

.694 

16 

1.777 

i        26 

4.694 

101 

.730 

161 

1  833 

!        26l 

4.876 

loi 

.766 

16i 

1.890 

27 

5.062 

loii 

.803 

16f 

1.948 

27I 

5.252 

11 

.840 

17 

2.006 

1        28 

5.444  1 

i4 

.878 

17i 

2.066 

i        28^ 

5.640 

.918 

17i 

2.126 

29 

5.840 

Hf 

.959 

17f 

2.187 

29j 

6.044 

' 

1.  Required   the   contents   of  a   piece   of  timber   whose 
length  is  30  feet  and  quarter  girt  is  17^  inches. 

Ans.  G5.G10  feet. 


64 


SELF-INSTRUCTOR 


Solution  hy  the  Table.  —  Look  for  the  quarter  girt  17|, 
in  the  oohmui  marked  (Quarter  Girt,  and  in  the  adjoining 
cohimn  marked  Area,  will  be  found  2.18G,  which  muhiplied 
by  the  length,  i30  feet,  will  be  65.610  feet  for  the  solid  eon- 
tents. 

Table  showing  the  Weight  in  Pounds  and  Decimals  of  a 
Pound  Avoirdupois  of  one  Cubic  Foot  of  the  following 
Kinds  of  Wood. 


i 


Cork  Wood 15.00 

Poi)lar 23.94 

Larch  or  Ilackmatiick  .     .  34,00 

Elm  and  West  India  Fir  .  34.75 

Maho;;any 35.00 

Pitch  Pine 41.25 

Cedar 37.25 

Pear  Tree 41.31 

Walnut 41.94 

Elder  Tree 43.44 

Beech 43.50 

Cherry  Tree 44.68 


Maple  and  Higa  Fir 
Ash  and  Dantzic  Oak 
Apple  Tree     .    .    . 

Alder 

Oak,  Canadian  .  . 
Boxwood,  French  . 
Logwood  .... 
Oak,  English  .  .  . 
Oak,  sixty  years  old 

Eb<>nj 

Lignum  Vitaj .    .     . 


46.87 
47.50 
49.56 
50.00 
54.50 
57.00 
57.06 
51.87 
73.12 
83.18 
83.31 


Ride  for  finding  the  Weight  of  any  hind  of  Timber. 

Multiply  the  number  of  cubic  feet  it  contains  by  the 
weight  of  one  cubic  foot  of  said  timber. 

EXAMPLES. 

1.  What  is  the  weight  of  a  piece  of  harkmn'ack  timber 
8  inches  X  1 2  inches,  and  30  feet  long  ? 

By  the   table   given  of   cubi         or  •       --hes  X  ^2 

inches  is  |  of  the  length,  for  t'  re  30  -j-  | 

=  20  feet,  contents. 

By  the  table  of  weight"^  k  able  fo  .  of  hackmatack  is 
=  to  34  lbs.,  therefore  34  X  ^^=  1.'  20  lbs.  avoirdupois. 


ON   LUMBER   SURVEYING. 


65 


46.87 

47.50 

49.56 

50.00 

54.50 

57.00 

57.06 

51.87 

73.12 

83.18 

«3.31 


X  12 


50 


2 
3 


2.  What  is  the  weij;ht  of  a  piece  of  Canadian  oak  12 
inelics  X  12  inches,  and  IK)  feet  long?        Ans.  l,G.'jr).()0  Ih.s. 

3.  What  is  the  weight  of  a  piece  of  Frencli  hoxwooil  10 
indies  X  12  inches,  and  24  feet  in  length? 

IJy  the  tahle  of  cnhic  measure,  10  inches  X  12  inches  is 
5  of  the  length,  for  the  contents  in  cubic  feet ;  therefore  24 
-^  5  =  20  feet,  contents;  20  X  ^7  =  1,140  lb.s.  =  weight 
required. 

P.  S.  —  The  weiglit  of  any  substance  may  be  found  as 
above,  by  finding  the  weight  of  1  cubic  f'X)t  and  multiplying 
said  weight  by  the  contents. 

TONNAGE  OF  VP:SSELS. 
Government  Rule.     English. 

For  vessels  aground,  the  length  is  to  be  measured  on  a 
straight  line  along  the  rabbet  of  the  keel,  from  a  ])erpendicu- 
lar,  let  full  from  the  back  of  the  main-post,  at  the  height  of 
the  wing-transom,  to  a  ])erpendicular  at  the  height  of  the 
upper  deck  (but  the  middle  deck  of  three-decked  ships), 
from  the  forepart  of  the  stern  ;  then  from  the  length  between 
these  perpendiculars  subtract  three  fifths  of  the  extreme 
breadth  for  the  rake  of  the  stern;  and  2},  inches  for  everv 
foot  of  the  height  of  the  wing-transom  above  the  lower  part 
of  the  rabbet  of  the  keel,  for  the  rake  abaft ;  and  the  re- 
mainder will  be  the  length  of  the  keel  for  tonnnge.  The 
main  breadth  is  to  be  taken  from  the  outside  of  the  outside 
plank,  in  the  broadest  part  of  the  ship  either  above  or  be- 
low the  wales,  deducting  therefrom  all  that  it  exceeds  the 
thickness  of  the  plank  of  the  bottom,  which  shall  be  ac- 
counted the  main  breadth  ;  so  that  the  moulding  breadth,  or 
the  breadth  of  the  frame,  Avill  then  be  less  than  the  main 
breadth,  so  found,  by  double  the  thickness  of  the  plank  of 
the  bottom. 

Rule.  —  Then  multiply  the  length  of  the  keel  for  tonnage, 
by  the  main  breadth,  so  taken,  and  the  product  by  half  the 


i  '\ 


66 


SELF-INSTRUCTOR 


breadth  ;  then  divide  the  whole  by  94,  and  the  quot'ient  will 
be  the  tonnage. 

In  cutters  and  brigs,  where  the  rake  of  the  stern-post  ex- 
ceeds 2?,  inches  to  every  foot  in  height,  the  actual  rake  is 
generally  subtracted  instead  of  the  2}j  inches  to  every  foot, 
as  before  mentioned. 

1.  Sujjpose  the  length  from  the  fore-part  of  the  stern,  at 
the  height  of  the  upper  deck,  to  the  after-part  of  the  stern- 
post,  at  the  height  of  the  wing-transom  to  be  115  feet  8 
inches^  the  breadth  from  outside  to  outside  40  feet  G  inches, 
and  the  height  of  the  wing-transom  21  feet  10  inches,  what  is 
the  tonnage?  Ans.  1,094. 

ft.  in. 

40   0  breadth 
3 

40  3X3=  120.9  ;  120.9  ^  5  =  ?A.l5. 

21.10  height  of  wing-transom  21.10  X  2^  =  04^7^  ;  54/^ 
-J-  12  =  4.55  ;  4.55  +  24.15  =  28.70 ;  155.66  —  28.70  = 
1 2  6.SG=  length. 

i2Gjjoxjo.p2o.i25  ^  1^094,  the  tonnage  required. 

2.  If  the  length  of  the  keel  be  120  feet,  and  the  breadth 
40  feet,  what  is  the  tonnage?  A7is.  1,02 l-j^  tons. 

Solution.  — 120  X  40  =  4,800  ;  4,800  X  20  =  96,000  ; 
96,000  -f-  94  =  1,021  if  tons. 

3.  If  the  length  of  the  keel  be  80  feet,  and  the  breadth 
of  the  beam  36  feet,  wliat  is  the  tonnage  ?  Ans,  55;|5. 

4.  If  the  length  of  the  keel  be  460  feet,  and  the  breadth 
of  the  beam  80  feet,  what  is  the  tonnage. 

Ans.  15,659  tons. 
Some  divide  the  last   product   by  100,  to  find  the  ton- 
nage of  king's  ships,  and  by  95,  to  find  that  of  merchant 
ships. 

Anncrican   Government  Rule. 

For  single-decked  i-essels.  —  Take  the  lengtli  on  deck  from 
the  forward  side  of  the  main  stern  to  the  after-side  of  the 
stern-post,  and  the  breadth  at  the  broadest  part  above  the 


mt  will 

)OSt  rx- 
rake  is 
iry  foot, 

item,  at 
e  stern- 
>  feet  8 
1  inches, 
what  is 
s.  1,094. 


28.70  = 


breadth 

[-\^  tons. 

9G,000  ; 

breadth 

breadth 

G59  tons, 
the  ton- 
merchant 


ON  LUMBER   SURVEYING. 


67 


ook  from 
do  of  the 
bovc  the 


main  wales  ;  take  the  depth  from  the  under  side  of  the  deck 
phuik  to  the  ceiling  of  the  hokl,  and  deduct  from  the  length 
tln-ee  fifths  of  the  breadth  ;  multiply  the  remainder  by  the 
breadth,  and  the  product  by  tlie  depth,  and  divide  the  last 
product  by  95. 

For  doiihle-deched  vessels.  —  Proceed  as  with  single-decked 
vessels,  except  for  the  depth  take  half  the  breadth. 

GAUGING. 

Gau";inij  sitriu'fies  the  art  of  measurin";  all  kinds  of  vessels 
and  determining  tlieir  capacity  or  the  quantity  of  fluid  or 
other  matter  tliey  contain.  It  is  usual  to  divide  casks  into 
four  varieties,  wliich  are  judged  of  from  the  greater  or  less 
apparent  curvature  of  their  sides,  namely  :  — 

1.  Tlie  middle  frustum  of  a  spheroid. 

2.  Tlie  middle  frustum  of  a  parabolic  spindle. 

3.  The  two  equal  frustums  of  a  paraboloid. 

4.  The  two  equal  frustums  of  a  cone. 

282  cubic  inches  make  1  ale  gallon,  or  beer. 
231  cubic  inches  make  1  wine  gallon. 
21,504  cubic  mches  make  1  malt  bushel. 

To  Jind  the  contents  of  a   Cask  by  the  Mean  Diameter. 

Rule.  — IMultiply  the  difference  of  the  head  and  bung  di- 
ameters by  .G8  for  the  first  variety ;  by  .02  for  tlie  second ; 
by  .55  for  the  third  ;  and  by  .5  for  the  fourth,  when  the  dif- 
ference between  the  head  and  bung  dinmeter  is  less  than  G 
inches ;  but  when  the  difference  between  these  exceeds  6 
inches,  multiply  that  difference  by  .7  for  the  first  variety  ;  by 
.G4  for  the  second;  by  .57  for  the  third;  and  by  .52  for  the 
fourth.  Add  this  product  to  the  head  diameter,  and  the 
sum  will  be  a  mean  diameter.  Square  this  mean  diameter, 
and  multiply  the  square  by  the  length  of  the  cask  ;  this 
product  multiplied  or  divided  by  the  proper  multiplier  or 
divisor,  will  give  the  contents. 

1.  What  are  the  contents  of  a  spheroidal  cask,  whose 


I 


68 


SELF-INSTRUCTOR 


length  is  40  inches,  bung  diameter  32  inches,  and  head  di- 
ameter 24  inches  ?  Ans.  97. G  gallons. 
Solution.- 32  —  24  =  8  ;  8  X  7  =  5.6  ;  5.0  -|-  24  ==  29.0 
=  mean  diameter;  29.0  X  29.0=  870.10  =  square  ;  870.10 
X  40  =  35040.40,  which  being  divided  by  359.5,  the  divisor 
for  imperial  gallons,  will  he  equal  to  97. G  gallons. 
By  the  gauging  rule  — 

Set  40  on  C.  to  the  G.  R.  18.79  on  D.  against    " 
24  on  D.  stands  04.99  on  C. 

32  on  D.  stands  110.2    on  C. 

+  110.2 


3)297.39 


99.13  gallons. 

Dr.  Iluttori's  General  Rule  for  fndiny  the  Contents  of  Casls. 

Add  into  one  sum  39  times  the  square  of  the  bung  diame- 
ter, 25  times  the  square  of  the  head  diameter,  and  20  times 
the  product  of  the  two  diameters  ;  then  multiply  the  sum  by 
the  length,  and  the  product  again  by. 00031^  for  the  con- 
tents in  gallons.         .  , 

EXAMPLE. 

1.  Whitt  are  the  contents  of  a  cask  whose  length  is  40 
inches,  and  the  bung  and  head  diameters  32  and  24  ? 

Ans.  93.4579  gallons. 
32  X  32  =  1024  ;  1024  X  39  =  39936 
24  X  24  =    570  ;     570  X  25  =  14400 
32X24=708;     708X20  =  19908 


74tJU4X  40  =  2972100 
.00031$ 


93.4579 
.^  Paging  is  the  art  of  finding  what  quantity  of  liquor  is 
contained  in  a  cask  when  partly  empty.     And  it  is  consid- 


ON   LUMBER   SURVEYING. 


69 


ead  (li- 

gullons. 

=  29.0 

87G.1G 

divisor 


f  Cash. 

r  diame- 

^6  times 

sum  by 

the  cou- 


th is  40 

V 
gallons. 


9721  GO 
.00031* 


3.4579 
liquor  is 
is  consid- 


ered in  two  positions  ;  first,  as  standing  on  its  end  ;  secondly, 
lying  on  its  side. 

To  Jiiul  the    Contents  of  Ullage  by  the  Sliding  Rule. 

By  one  of  the  preceding  problems  find  the  whole  con- 
tents of  the  cask.  Then  set  the  length  on  N.  to  100  on  S. 
S.  for  a  segment  standing,  or  set  the  bung  diameter  on  N.  to 
100  on  S.  L.  for  a  segment  lying;  then  against  the  wet 
inciies  on  N.  is  a  number  on  S.  S.  or  S.  L.  to  be  reserved. 
Next  set  100  on  B.  to  tiie  reserved  number  on  A. ;  then 
against  the  whole  contents  on  B.  will  be  found  the  ullage 
on  A. 

QUESTIONS    FOR    EXERCISE. 

1.  What  are  the  contents  of  20  pieces  of  timber  8  inches 
X  12  inches,  and  36  feet  long  in  cubic  feet,  and  also  in  su- 
perficial feet  ? 

2.  What  number  of  cubic  feet  in  a  log  whose  quarter  girt 
is  17|-  irshes  and  length  18  feet  ? 

3.  What  are  the  contents  of  24  logs  16  feet  louff  whose 
quarter  girt  is  27  inches  ? 

4.  Required  the  tonnage  of  a  ship  by  the  English  and 
American  rules,  the  length  of  the  keel  beini;  125  feet  and 
the  breadth  of  the  beam  42  feet  ? 

5.  What  is  the  weiglit  of  a  })iece  of  hackmatack  timber  8 
inches  X  10  inches  and  28  feet  in  lenf>-th  ? 

6.  Required  the  number  of  tons  in  16  pieces  of  timber 
24  feet  long  and  12  inches  X  16  inches? 

7.  In  2,500  feet  running  lengtii  of  2  inches  X  10  inches, 
how  many  feet  of  board  measure  ? 

8.  In  300  feet  ruiming  length  of  10  hich  X  12  inch  tim- 
ber, how  many  tons  ? 

9.  Wliat  are  the  contents  of  a  cask  of  the  first  variety  in 
■wine  and  ale  gallons,  whose  length  is  50  inches,  bung  diam- 
eter 38  inches,  and  head  diameter  30  inches  ? 

10.  If  a  log  be  35  indies  in  diameter,  what  is  the  larsfest 
piece  of  square  timber  that  can  be  sawed  from  it  ? 


i 


70 


SELF-INSTRUCTOR 


11.  What  difference  is  there  between  a  floor  28  feet  long 
X  20  feet  broad,  and  two  others,  each  of  half  the  dimen- 
sions ;  and  what  do  the  three  floors  come  to  @  S9.00  per 
100  square  feet  ?  Ans.  $75.'60. 

12.  An  elm  plank  is  14  feet  3  inches  long,  and  it  is  de- 
sired that  just  a  square  yard  may  be  slit  ofl'from  it;  at  what 
distance  from  the  edge  must  the  line  be  struck  ? 

Ans.  7^^^^  inches. 

13.  A  joist  is  7  inches  wide  and  2^  inches  thick,  but  a 
scantling  just  as  big  again,  that  shall  be  3  niches  thick,  is 
wanted  ;  what  will. the  other  dimension  be? 

Ans.  11§  inches. 

14.  Tlie  perambulator  is  so  contrived  as  to  tin-n  just 
twice  in  IG^  feet;  required  the  diameter?     Ans.  2.G2G  feet. 

15.  In  turning  a  chaise  witlun  a  ring  of  a  certain  diameter, 
it  was  observed  that  the  outer  wheel  made  two  revolutions 
while  the  inner  made  but  one  ;  the  wheels  were  both  4  feet 
high,  and  supposing  them  fixed  at  the  distance  of  5  feet 
asunder  on  the  axletree,  wliat  was  the  circumference  of  the 
track  described  by  the  outside  wheel  ?     Ans.  G3  feet  nearly. 

IG.  Having  a  rectangular  board  58  inches  by  27  inches, 
I  would  have  a  square  foot  cut  off  parallel  to  the  shorter 
edge ;  I  would  then  have  the  same  quantity  cut  from  the 
remainder,  parallel  to  the  longer,  and  this  alternately  re- 
peated, till  there  shall  not  be  the  quantity  of  a  foot  left ; 
what  will  be  the  dimensions  of  the  remaining  piece  ? 

A?is.  20.7  inches  by  6.086. 

17.  What  is  the  lenj»th  of  a  chord  which  cuts  off  ^  of 
the  area  of  a  circle,  whose  diameter  is  289  ? 

Ans.  278.G716. 

18.  What  Avill  the  diameter  of  a  globe  be,  when  the  solid- 
ity an  !  superficial  contents  are  expressed  by  the  same  num- 
ber ?  Ans.  6. 

19.  A  gentleman  has  a  garden  100  feet  long  and  80  feet   . 
broad,  and  a  gravel  walk  is  to  be  made  of  an  equal  width 
half  round  it ;  what  must  be  the  breadth  of  the  walk  to 
take  up  just  half  the  ground?  Ans  25.9G8  feet. 


ON  LU  i3ER   SURVEYING. 


n 


,'  6.086. 


Ans. 


20.  How  many  3-inch  cubes  may  be  cut  out  of  a  1 2-inch 
cube?  Alls.  64:. 

21.  How  high  above  the  earth  must  a  person  be  raised 
that  he  may  see  one  third  of  its  surface  ? 

Ans.  To  the  lieight  of  the  earth's  diameter. 

22.  How  many  feet  of  boards  would  cover  tlie  surface  of 
the  earth,  its  diameter  being  7,958  miles ;  and  how  many 
solid  feet  in  it  ?  ' 

f  5,546,407,680,000,000.     No.  of 
feet  of  boards  to  cover  it. 
37,416,291,092,323,844,085,000. 
No.  of  cubic  feet  in  the  earth. 

23.  If  the  diameter  of  a  circle  be  50  feet,  what  is  the 
circumference  of  it  ? 

24.  Two  pillars  standing  on  a  horizontal  plane  are  120 
feet  asunder;  the  height  of  the  higher  is  100  feet,  and  tliat 
of  the  lower  80 ;  whereabout  in  tiie  plane  must  a  person 
place  himself,  so  that  his  distance  from  tlie  top  of  eitlier  of 
the  piliars  shall  be  equal  to  the  distance  between  them  ? 

Ans.  91.78  feet  from  the  bottom  of  the  lower, 
69.92  feet  from  the  bottom  of  the  other. 

25.  Three  ships  are  equally  distant  from  an  island,  the 
first  shin  is  30  miles  from  tlie  second,  the  second  is  25  miles 
from  the  third,  and  the  third  is  20  miles  from  the  first ;  re- 
quired the  distance  to  the  isle  ? 

An.',.  15.118579  miles  from  each. 

26.  Prove  that  the  elevation  of  the  North  or  Polar  star 
above  the  horizon  is  equal  to  the  latitude  of  the  place  where 
its  altitude  is  taken. 

27.  I  have  a  board  in  the  form  of  a  triangle ;  the  length 
of  one  of  its  sides  is  16  feet.  I  wish  to  sell  one  half  of  it; 
at  what  distance  from  the  larger  end  must  it  be  divided  par- 
allel to  the  larger  end.  Ans.  4.68  feet. 

28.  In  2,500  feet  running  lengths  of  7  inches  X  ^  inches, 
how  many  feet  running  lengths  of  2^  inches  X  1 1  ? 


72 


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—  p^  —  —  —  p_  rt  —  —  —  71  71  71  71  71   71  71  71  (M  G^ 


ON   LUMBER   SURVKYING. 


77 


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Hoio  to  use  the  Log  or  Timber  Rule, 

If  tlic  timber  i.s  taI)erin^^  the  girt  should  be  taken  nbout 
one  third  the  di.stsince  from  the  hirger  to  the  .sinalU'r  end. 
Some  take  the  <,nrt  in  the  middle.  Girt  the  lof;  to  be  meas- 
ured, and  take  the  quarter  of  it,  and  measure  the  len<!;th  of 
the  log.  Then  look  along  the  to))  of  the  table  till  yon  come 
to  the  corresponding  (jnarter  girt;  then  rundown  the  colunm 
underneath  the  quarter  girt  till  you  get  opposite  the  length, 
Avhere  you  will  tind  the  contents.  Or,  you  can  iind  the  con- 
tents by  taking  the  diameter  of  the  small  end  and  (he  length. 
Then  find  the  corresponding  diameter  at  the  foot  of  the  table, 
and  ascend  the  line  perpendicularly  till  you  come  opposite 
the  length,  where  you  will  find  the  contents. 

!*•  S.  —  This  table  allows  one  fourth  of  the  true  contents 
of  the  log  for  bark,  saw  kerf,  and  waste  slab.  It  has  been 
extensively  used  by  timber  merchants,  and  is  just  about  as 
fair  a  rule  to  go  by  as  any  I  have  seen.  There  are  many 
allowances  to  be  made  which  are  left  to  the  scaler's  iud<»-- 
ment,  and  for  wliich  it  would  be  almost  impossible  to  make 
due  allowance  in  the  tabic. 


INTEREST. 


Rule  for  Jill  ding  the  interest  at  G  per  cent.  —  Multiply  the 
sum  by  the  number  of  days,  divide  the  product  by  G,  then 


strike  off  the  right-hand  figure. 


EXAMPLE. 

$200 
1 2  days. 


G)2400 


400  =  40  cents  is  the  interest. 


mmm 


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Hrs<  Examjjle  at  Q  per  cen^  — ■Rpquired,  50  days  interest  on  $100. 
Interest  on  $100  for  30  days  =  49  cents. 
Interest  on  $100  for  20  days  =  33     " 


Am.  82  cents. 


INTKUKST. 


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Second  Example  at  &  per  cent.  —  Required,  the  interest  of  $50  for  3  years, 
2  months,  and  10  days. 

Interest  on  S50  for  .?  years  =  $9.00 
Interest  on  S50  for  2  mos.  r=  50 
Interest  on  $50  for  10  days  =        8 


Ans.     $!).C8 


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WANTED. 

Agents  to  sell  this  Book,  throughout  the  United  States, 
the  Dominion  of  Canada,  California,  and  Oregon.  Exchi- 
sive  territory  given.  Good  inducements  to  agents.  The 
book  will  be  sent  to  any  address,  free  of  postage,  on  receipt 
of  Two  Dollars.  Send  Po.st-Office  orders,  or  by  express. 
Address 

CHARLES   KINSLEY, 
Calais,  Ale.,  or  St.  Stephen,  N.  B. 


NOTE. 

All  Lumber  Manufacturers,  Lumber  Dealers,  Millmen, 
Carpenters,  Carriage  jVIakers,  Shipbuilders,  Cabinet  Mak- 
et-s,  Ship  Brokers,  Ship  Carpenters,  Riilroad  Conductors, 
Engineers,  Machinists,  Freight  Agents,  Teachers,  Students, 
Architects,  Merchants,  Accountants,  and  others,  will  find 
it  to  their  advantage  to  procure  a  copy  of  this  book,  as  the 
knovdedge  it  imparts  may  save  them  in  a  few  years'  prac- 
tice hundreds  of  dollars.  The  book  contains  twelve  new 
rules  for  finding  the  superficial  contents  of  lumber,  which 
do  the  same  work  as  one  hundred  and  fifty  of  the  rules 
generally  used. 

CHARLES   KINSLEY. 


states, 
Exclu- 
The 
Gceipt 
ress. 

f , 

LB. 


llmen, 
;  Mak- 
uctors, 
idents, 
11  find 

as  the 
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which 
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EY. 


